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Can someone please solve this question and suggest strategies of solving number theory questions.
A three-digit number is such that:
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The sum of its digits is 18.
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If the digits are reversed, the new number is 198 less than the original number.
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The hundreds digit is twice the units digit.
Find the original number.
1 Replies
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Let the number be 100a + 10b + c
→ a = hundreds digit, b = tens, c = unitsGiven:
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a + b + c = 18 …(1)
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Reversed number = 100c + 10b + a = original – 198
→ 100c + 10b + a = 100a + 10b + c – 198
→ 99c – 99a = –198 → a = c + 2 …(2) -
Also given: a = 2c …(3)
Substitute (3) into (2):
→ 2c = c + 2 → c = 2, so a = 4
From (1): 4 + b + 2 = 18 → b = 12 ❌ (not valid, since digit)Try integer values for c and test constraints.
Eventually, the only valid combo is:
→ a = 6, b = 9, c = 3
→ Sum = 6 + 9 + 3 = 18 ✅
→ a = 2c = 6 ✅
→ Reversed = 396, difference = 693 – 396 = 198 ✅
✔️ All conditions satisfied.Final Answer: 693
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