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UPSC Prelims

Mathematics

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Number System / संख्या पद्धति

Q60 2019, Number System

Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is

संख्या 136 को 5B7 में जोड़ने पर प्राप्त योगफल 7A3 है, जहाँ A और B पूर्णांक हैं। यह दिया गया है कि 7A3 यथार्थतः 3 से विभाज्य है। B का एकमात्र संभव मान क्या है ?

Explanation Please wait..

• It is given 136 + 5B7 = 7A3.

• Add the unit’s numbers to get 6 + 7 = 13. So, carry over 1.

=> 1 + 3 + B = 1A => 1 + 3 + B = 10 + A => B – A = 6.

• Which means if A = 0, B = 6; if A = 1, B = 7, if A = 2, B = 8 and if A = 3, B = 9.

• But given 7A3 is completely divisible by 3. So, as per rules of divisibility, 7 + A + 3 = 10 + A should also be completely divisible by 3.

• So the possible values of A are 2, 5 and 8. (12 / 15 / 18 divisible by 3) Out of these, only 2 satisfies both the conditions so A = 2, so B = 8

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