1> If 84^n divides (496)! , then what is the maximum value of n ??
2> What is the number of 5 digit multiples of 3 that end with digit 6 ??
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5 Answers
62496 = 124x4 = 31x16 = 31x24
84 = 3x7x22
highest power of 7 in 496 is 81
Highest power of 3 is not the limiting point..
while the highest power of 2 in 496 is 491 (check this one once)
so the limiting power is that of 7
so the highest value of n is 81
49this is a simple problem to be done with the motivation of finding the no of terms of am AP with first term, last term and common difference!!!!!!wat we r supposed is to logic out the first term last term and the cd
1Agree with subhomoy yet again!
We observe that for divisibility by 3,the first 5 digit no. ending with 6 is 10026 and last term will be 99996 and common diff of AP=30.
99996=10026+(n-1).30
=>n-1=2999
therefore n=3000.