{"id":90170,"date":"2025-06-01T10:22:12","date_gmt":"2025-06-01T10:22:12","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90170"},"modified":"2025-06-01T10:22:12","modified_gmt":"2025-06-01T10:22:12","slug":"a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/","title":{"rendered":"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If"},"content":{"rendered":"

A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If 13Y7 is divisible by 11, then what is the value of (X+Y) ?<\/p>\n

[amp_mcq option1=”15″ option2=”12″ option3=”11″ option4=”10″ correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe addition is 4X3 + 984 = 13Y7.
\nLet’s perform the addition column by column:
\nUnits place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place.
\nTens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10.
\nHundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place.
\nSo, X + 8 = Y, where Y is a digit between 0 and 9.
\nThe possible values for X (a digit between 0 and 9) that result in a single digit Y are:
\nIf X=0, Y = 0+8 = 8. The sum is 1387.
\nIf X=1, Y = 1+8 = 9. The sum is 1397.
\nIf X=2, Y = 2+8 = 10. Y cannot be 10 as it’s a single digit.
\nSo, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397).<\/p>\n

We are given that 13Y7 is divisible by 11.
\nUsing the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11.
\nFor 13Y7: +7 – Y + 3 – 1 = 9 – Y.
\nFor 13Y7 to be divisible by 11, 9 – Y must be a multiple of 11 (0, 11, -11, etc.).
\nSince Y is a digit (0-9), 9 – Y can range from 9-0=9 to 9-9=0.
\nThe only multiple of 11 in this range is 0.
\nSo, 9 – Y = 0, which means Y = 9.<\/p>\n

Now we use the relation X + 8 = Y. Substitute Y=9:
\nX + 8 = 9
\nX = 9 – 8 = 1.
\nSo, X=1 and Y=9.<\/p>\n

Let’s verify: 413 + 984 = 1397.
\nIs 1397 divisible by 11? 7 – 9 + 3 – 1 = 0. Yes, it is.<\/p>\n

The value requested is (X + Y).
\nX + Y = 1 + 9 = 10.
\n<\/section>\n

\nThis problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.
\n<\/section>\n
\nThe divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of the digits at odd positions (from the right) and the sum of the digits at even positions (from the right) is divisible by 11 (i.e., 0, 11, -11, 22, etc.).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If 13Y7 is divisible by 11, then what is the value of (X+Y) ? [amp_mcq option1=”15″ option2=”12″ option3=”11″ option4=”10″ correct=”option4″] This question was previously asked in UPSC CAPF – 2017 Download PDFAttempt Online The addition is 4X3 + 984 … <\/p>\n

Detailed SolutionA 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1085],"tags":[1101,1102],"class_list":["post-90170","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1101","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nA 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If<\/title>\n<meta name=\"description\" content=\"The addition is 4X3 + 984 = 13Y7. Let's perform the addition column by column: Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place. Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10. Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place. So, X + 8 = Y, where Y is a digit between 0 and 9. The possible values for X (a digit between 0 and 9) that result in a single digit Y are: If X=0, Y = 0+8 = 8. The sum is 1387. If X=1, Y = 1+8 = 9. The sum is 1397. If X=2, Y = 2+8 = 10. Y cannot be 10 as it's a single digit. So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397). We are given that 13Y7 is divisible by 11. Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11. For 13Y7: +7 - Y + 3 - 1 = 9 - Y. For 13Y7 to be divisible by 11, 9 - Y must be a multiple of 11 (0, 11, -11, etc.). Since Y is a digit (0-9), 9 - Y can range from 9-0=9 to 9-9=0. The only multiple of 11 in this range is 0. So, 9 - Y = 0, which means Y = 9. Now we use the relation X + 8 = Y. Substitute Y=9: X + 8 = 9 X = 9 - 8 = 1. So, X=1 and Y=9. Let's verify: 413 + 984 = 1397. Is 1397 divisible by 11? 7 - 9 + 3 - 1 = 0. Yes, it is. The value requested is (X + Y). X + Y = 1 + 9 = 10. This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If\" \/>\n<meta property=\"og:description\" content=\"The addition is 4X3 + 984 = 13Y7. Let's perform the addition column by column: Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place. Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10. Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place. So, X + 8 = Y, where Y is a digit between 0 and 9. The possible values for X (a digit between 0 and 9) that result in a single digit Y are: If X=0, Y = 0+8 = 8. The sum is 1387. If X=1, Y = 1+8 = 9. The sum is 1397. If X=2, Y = 2+8 = 10. Y cannot be 10 as it's a single digit. So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397). We are given that 13Y7 is divisible by 11. Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11. For 13Y7: +7 - Y + 3 - 1 = 9 - Y. For 13Y7 to be divisible by 11, 9 - Y must be a multiple of 11 (0, 11, -11, etc.). Since Y is a digit (0-9), 9 - Y can range from 9-0=9 to 9-9=0. The only multiple of 11 in this range is 0. So, 9 - Y = 0, which means Y = 9. Now we use the relation X + 8 = Y. Substitute Y=9: X + 8 = 9 X = 9 - 8 = 1. So, X=1 and Y=9. Let's verify: 413 + 984 = 1397. Is 1397 divisible by 11? 7 - 9 + 3 - 1 = 0. Yes, it is. The value requested is (X + Y). X + Y = 1 + 9 = 10. This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:22:12+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If","description":"The addition is 4X3 + 984 = 13Y7. Let's perform the addition column by column: Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place. Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10. Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place. So, X + 8 = Y, where Y is a digit between 0 and 9. The possible values for X (a digit between 0 and 9) that result in a single digit Y are: If X=0, Y = 0+8 = 8. The sum is 1387. If X=1, Y = 1+8 = 9. The sum is 1397. If X=2, Y = 2+8 = 10. Y cannot be 10 as it's a single digit. So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397). We are given that 13Y7 is divisible by 11. Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11. For 13Y7: +7 - Y + 3 - 1 = 9 - Y. For 13Y7 to be divisible by 11, 9 - Y must be a multiple of 11 (0, 11, -11, etc.). Since Y is a digit (0-9), 9 - Y can range from 9-0=9 to 9-9=0. The only multiple of 11 in this range is 0. So, 9 - Y = 0, which means Y = 9. Now we use the relation X + 8 = Y. Substitute Y=9: X + 8 = 9 X = 9 - 8 = 1. So, X=1 and Y=9. Let's verify: 413 + 984 = 1397. Is 1397 divisible by 11? 7 - 9 + 3 - 1 = 0. Yes, it is. The value requested is (X + Y). X + Y = 1 + 9 = 10. This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/","og_locale":"en_US","og_type":"article","og_title":"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If","og_description":"The addition is 4X3 + 984 = 13Y7. Let's perform the addition column by column: Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place. Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10. Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place. So, X + 8 = Y, where Y is a digit between 0 and 9. The possible values for X (a digit between 0 and 9) that result in a single digit Y are: If X=0, Y = 0+8 = 8. The sum is 1387. If X=1, Y = 1+8 = 9. The sum is 1397. If X=2, Y = 2+8 = 10. Y cannot be 10 as it's a single digit. So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397). We are given that 13Y7 is divisible by 11. Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11. For 13Y7: +7 - Y + 3 - 1 = 9 - Y. For 13Y7 to be divisible by 11, 9 - Y must be a multiple of 11 (0, 11, -11, etc.). Since Y is a digit (0-9), 9 - Y can range from 9-0=9 to 9-9=0. The only multiple of 11 in this range is 0. So, 9 - Y = 0, which means Y = 9. Now we use the relation X + 8 = Y. Substitute Y=9: X + 8 = 9 X = 9 - 8 = 1. So, X=1 and Y=9. Let's verify: 413 + 984 = 1397. Is 1397 divisible by 11? 7 - 9 + 3 - 1 = 0. Yes, it is. The value requested is (X + Y). X + Y = 1 + 9 = 10. This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:22:12+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/","name":"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:22:12+00:00","dateModified":"2025-06-01T10:22:12+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The addition is 4X3 + 984 = 13Y7. Let's perform the addition column by column: Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place. Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10. Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place. So, X + 8 = Y, where Y is a digit between 0 and 9. The possible values for X (a digit between 0 and 9) that result in a single digit Y are: If X=0, Y = 0+8 = 8. The sum is 1387. If X=1, Y = 1+8 = 9. The sum is 1397. If X=2, Y = 2+8 = 10. Y cannot be 10 as it's a single digit. So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397). We are given that 13Y7 is divisible by 11. Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11. For 13Y7: +7 - Y + 3 - 1 = 9 - Y. For 13Y7 to be divisible by 11, 9 - Y must be a multiple of 11 (0, 11, -11, etc.). Since Y is a digit (0-9), 9 - Y can range from 9-0=9 to 9-9=0. The only multiple of 11 in this range is 0. So, 9 - Y = 0, which means Y = 9. Now we use the relation X + 8 = Y. Substitute Y=9: X + 8 = 9 X = 9 - 8 = 1. So, X=1 and Y=9. Let's verify: 413 + 984 = 1397. Is 1397 divisible by 11? 7 - 9 + 3 - 1 = 0. Yes, it is. The value requested is (X + Y). X + Y = 1 + 9 = 10. This problem combines basic arithmetic addition with the divisibility rule for 11 and digit constraints.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-3-digit-number-4x3-is-added-to-984-to-get-a-4-digit-number-13y7-if\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90170","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=90170"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90170\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90170"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90170"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90170"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}