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)

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Submitted by examrobotsa on

Q: 4 (CAPF/2023)
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)

question_subject: 

Mathematics

question_exam: 

CAPF

Given the equation (100a + 10b + c) - (100c + 10b + a) = 396, simplifying yields 99(a - c) = 396. This simplifies to a - c = 4. Given the condition of non-repeating digits and reasonable ranges for a, b, and c, only 4 such numbers satisfy these constraints as valid three-digit numbers.