There are how many integers "n" in between 1 and 10^7, including both, such that last 7 digits of n and n^3 are
the same? (If n or n^3 has fewer than 7 digits,
then add 0s in the left and then compare the last 7 digits.)
a. 10 b. 15 c. 20 d.
None of These
Sol [b]
If last 7 digits of n and n^3 are the same it means n^3 - n
is divisible 10^7
i.e (n-1)*n*(n+1) is divisible by 10^7 [n^3 - n=(n-1)*n*(n+1)]
It shows (n-1)*n*(n+1) is divisible by both 2^7 & 5^y
Case I: (n-1)*n*(n+1) is divisible by 2^7
Sub-case a : If n is even than both (n-1) & (n+1)are odd so n should be
divisible by 2^7
=> n is multiple of 2^7.......................(i)
=>n = 2^7, 2*2^7,.........,5^7*2^7
Sub-case b : If n is odd than (n-1)*(n+1) should be
divisible by 2^7
Since the gap between n-1 and n+1 is 2 so one of them is divisible by 4 and
other is by only 2
Sub-case b1 : Let n-1 is divisible by 2 but 4 , then n+1 is
divisible by 2^6 => (n+1) is multiple of 2^6
=> n = 2^6 -1 , 2*2^6 -1, 5^7*2^6 -1 .....5^7*2*2^6 - 1
Sub-case b2: Let n+1 is divisible by 2 but 4 , then n-1 is
divisible by 2^6 => (n-1) is multiple of 2^6
=> n = 1, 2^6 +1 , 2*2^6 +1, 5^7*2^6 +1 .....
Sub-case b3 : Let n-1 is divisible by 2^7 => (n-1)
is multiple of 2^7
Sub-case b4 : Let n+1 is divisible by 2^7 => (n+1)
is multiple of 2^7
So we
can say we are getting 5 different remainders for "n" when divided by
2^7 which are 0, 1, 2^6-1, 2^6+1 & 2^7 -1
Case II: (n-1)*n*(n+1) is divisible by 5^7
Sub-case (i) : n-1 divisible by 5^7 => n = k*5^7 +1
Sub-case (ii) : n divisible by 5^7 => n = k*5^7
Sub-case (iii) : n+1 divisible by 5^7 => n = k*5^7 -1
So we
can say we are getting 3 different remainders for "n" when divided by
5^7 which are 0, 1, -1
There are three possible values of remainders by 5^7 and five possible values by
2^7 . Then by the CRT (Chinese Remainder Theorem) there are 15 possible values
of remainder by 10^7.
Hence there are 15 required solutions of n in between 1 to 10^7.
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