How many three-digit numbers are possible such that the difference between the original number and the number obtained by reversing the digits is 396? (no digit is repeated)<\/p>\n
[amp_mcq option1=”4″ option2=”5″ option3=”50″ option4=”40″ correct=”option4″]<\/p>\n
If we strictly follow N – N_rev = 396, there are 6 * 8 = 48 such numbers. However, 48 is not among the options.<\/p>\n
Let’s consider a common convention in such problems: the reversed number must also be a three-digit number. This implies that the units digit of the original number, c, cannot be 0. How many three-digit numbers are possible such that the difference between the original number and the number obtained by reversing the digits is 396? (no digit is repeated) [amp_mcq option1=”4″ option2=”5″ option3=”50″ option4=”40″ correct=”option4″] This question was previously asked in UPSC CAPF – 2023 Download PDFAttempt Online Let the three-digit number be 100a + 10b … <\/p>\n
\nIf c must be in {1..9} (and a in {1..9}) with a-c=4:
\n– If c = 1, a = 5. Pair (5, 1).
\n– If c = 2, a = 6. Pair (6, 2).
\n– If c = 3, a = 7. Pair (7, 3).
\n– If c = 4, a = 8. Pair (8, 4).
\n– If c = 5, a = 9. Pair (9, 5).
\nThere are 5 such pairs for (a, c) if c!=0.
\nFor each of these 5 pairs, b must be distinct from a and c. There are 10 – 2 = 8 possible digits for b.
\nTotal number of such three-digit numbers = 5 pairs * 8 options for b per pair = 40.
\nThis matches option D. This suggests the implicit condition that the reversed number is also a three-digit number (c!=0) was intended.
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