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Answered by
15
Consider the number be x
when,
and so on to
Now , go to back studies we have a statement
If the smallest number which when divided by p,q and r leaves remainder x,y and z where (p-x) = ( q-y) = (r-z) = constant then the number will be LCM(p,q,r) - constant
here ,
2-1 = 1 , 3-2 = 1 , 4-3 = 1 to 10-9 = 1
We get as a constant.
So
x = (2×2×2×3×3×5×7)-1
x = 2,520 -1
x = 2,519
The number is
Answered by
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● Hope this answer will help u
N + 1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9 & 10 so clearly one solution would be 10! (ie 10*9*8*7*6*5*4*3*2 = 3628800 for N +1) but this is not the lowest possible. Follow the logic below...
N + 1 must be a multiple of 2
N + 1 must be a multiple of 4 but if it is a multiple of 2 & 4 it is necessarily a multiple of 8
N + 1 must be a multiple of 5 but if it is a multiple of 2 & 5 it is necessarily a multiple of 10
N + 1 must be a multiple of 6 but if it is a multiple of 2 & 9 (18) it is necessarily a multiple of 6
N + 1 must be a multiple of 7
N + 1 must be a multiple of 9
Hence 2 x 4 x 5 x 7 x 9 =
2520 is a multiple of 2, 3, 4, 5, 6, 7, 8, 9 & 10
N + 1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9 & 10 so clearly one solution would be 10! (ie 10*9*8*7*6*5*4*3*2 = 3628800 for N +1) but this is not the lowest possible. Follow the logic below...
N + 1 must be a multiple of 2
N + 1 must be a multiple of 4 but if it is a multiple of 2 & 4 it is necessarily a multiple of 8
N + 1 must be a multiple of 5 but if it is a multiple of 2 & 5 it is necessarily a multiple of 10
N + 1 must be a multiple of 6 but if it is a multiple of 2 & 9 (18) it is necessarily a multiple of 6
N + 1 must be a multiple of 7
N + 1 must be a multiple of 9
Hence 2 x 4 x 5 x 7 x 9 =
2520 is a multiple of 2, 3, 4, 5, 6, 7, 8, 9 & 10
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