Math, asked by science4128, 9 months ago

What is the least no which when divided by 52 leaves 33 as remainder and then divided by 78 leaves 59 when divided by 117 leaves 98

Answers

Answered by Shailesh183816
1

\bf\large\underline\blue{Solution:-}

➤ Let number be X. 

So X= 52N+33=78M+59=117P+98, where M,N, P are divisors(?? I guess). 

We know that the smallest number which can be divided by 52, 78 and 117 is 13x2x2x3x3 = 52x9 = 468. 

Now, look at the remainders in each case, I can write 33 as (52-19), 59 as 78-19, and 98 as 117-19. Did you find a pattern. yes. 

Now rewrite the equation. 

X = 52(N+1) - 19 = 78(M+1) - 19 and 117(P+1) - 19. 

So now, the ques. boils down to find GCD of 52,78,and 117 ( which we know = 468 ) and deduct 19 from it = 449. 

Just to test the answer 449 / 52 = 9 + 33 remainder. 

449/78 = 5 + 59 remainder, and 449/117 = 3 + 98 remainder. 

So Answer is 449. 

2. A. It gains time. Instead coinciding every 65+5/11 mins. hands coincide every 65 mins. 

so gain of 5/11 every 65 mins.

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