Question
A three-digit number is such that the sum of its digits
is 11. If you reverse its digits, the new number is 297 more than the original number. Find the original number.Solution
ATQ, Let the number be 100a + 10b + c, where a is hundreds digit, c is units digit. Given: a + b + c = 11 …(1) Reversed number = 100c + 10b + a Given: (100c + 10b + a) − (100a + 10b + c) = 297 ⇒ 100c + 10b + a − 100a − 10b − c = 297 ⇒ 99c − 99a = 297 ⇒ 99(c − a) = 297 ⇒ c − a = 3 …(2) From (1): a + b + c = 11 From (2): c = a + 3 Substitute: a + b + (a + 3) = 11 2a + b + 3 = 11 2a + b = 8 …(3) Digits 0–9, and a ≠ 0. Try integer values: From (2), c = a + 3 ≤ 9 ⇒ a ≤ 6 Test a = 2 ⇒ c = 5, then (3): 2×2 + b = 8 ⇒ b = 4 Number = 245 Check reversed 542 − 245 = 297 (correct) Sum of digits = 2 + 4 + 5 = 11 (correct). Answer: 245.
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31, 93, 279
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A = (5/2) × B and C = (1/10) × A. If (A + B + C) = 255, then determine the value of (7C – B).
Determine the number of trailing zeros in the product of the following numbers:
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Find the difference between minimum and maximum value of 'k' such that '6k8194' is always divisible by 3.
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