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Given a positive integer N and two of its divisors. Difference between N and these two divisors is 270 and 280...

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We are given a positive integer N. Two of its positive divisors are chosen and the differences between N and these two divisors are 270 and 280 respectively.

Find the number of possible value(s) of N?

posted Jun 28 by anonymous

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1 Answer

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If 'a', 'b' are the 2 positive divisors of N & if (N - a) = 270 & (N - b) = 280, then a > b or in other words a = b + 10
a = N - 270 & b = N - 280, which means N has to start from 280 for 'a' & 'b' to be positive.

285(N) is divisible by 15(a) & 5(b) ====> 15*19 & 5*57 ====> LCM = 15 < 285
288(N) is divisible by 18(a) & 8(b) ====> 18*16 & 8*36 ====> LCM = 72 < 288
300(N) is divisible by 30(a) & 20(b) ====> 30*10 & 20*30 ====> LCM = 60 < 300
315(N) is divisible by 45(a) & 35(b) ====> 45*7 & 35*9 ====> LCM = 315 = 315
are 4 instances for which the above conditions held true. This is partly because as 'a' and 'b' becomes bigger compared to 'N', 'a' and 'b' stop having an LCM lesser than N. In other words 'a' & 'b' will not divide 'N' simultaneously.

answer Nov 7 by Tejas Naik



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