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Re: Find the sum of last three digits of S, where S= 871*873*875*878*881*8 [#permalink]
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1. When u divide any number by 1000, remainder is last 3 digits of that number.
2. Remainder of X*Y, divided by 100= Remainder of X when divided by 100 * Remainder of Y when divided by 100

Find the Remainder of (871 * 873 * 7 * 439 * 881 * 883), when divided by 4
Remainder of 871, when divided by 4=3
Remainder of 873, when divided by 4=1
Remainder of 7, when divided by 4=3
Remainder of 439, when divided by 4=3
Remainder of 881, when divided by 4=1
Remainder of 883, when divided by 4=3

the Remainder of (871 * 873 * 7 * 439 * 881 * 883)= Remainder of 3*1*3*3*1*3=81, when divided by 4= 1
Hence we can write it as 4k+1

S/250= 4k+1
S=1000k +250
Now divide S by 1000, you will get remainder 250, which is our last 3 digits



sach24x7 wrote:
nick1816 wrote:
S= 871 * 873 * 875 * 878 * 881 * 883
S/250 = (871 * 873 * 875 * 878 * 881 * 883)/250
S/250= 871 * 873 * 7 * 439 * 881 * 883

871 * 873 * 7 * 439 * 881 * 883 ≡ 1 (mod 4)
Hence we can write 871 * 873 * 7 * 439 * 881 * 883 as 4k+1 .
S/250 =4k+1
S=1000k+250
Hence last 3 digits of S is 250.
Sum of last 3 digits 2+5+0=7


Hi

Could not understand the method you used?
Can you please explain?

Posted from my mobile device
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Re: Find the sum of last three digits of S, where S= 871*873*875*878*881*8 [#permalink]
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S= 871 * 873 * 875 * 878 * 881 * 883

Out of this we can see that there is one even number and one multiple of 5:
875 Factorises to 5 * 175 = 5^3 X 7
878 = 2 * 439

So S = 871 * 873 * (5^3) * 7 * 2* 439 *881 *883
But notice that (5^3)*2 = 250. This means S is a multiple of 250
S = 250 * (871 * 873 * 7 * 439 * 881 * 883)
S/250 = 871 * 873 * 7 * 439 * 881 * 883

So potential endings for S are 000, 250, 500 or 750

We know that there are no other multiples of 2 so that means it can't end in 000 or 500. It also means that there will be a remainder when divided by 4.

Then as mentioned above, the remainders of products = product of remainders.

Let N = S/250
N = 4M + Remainder
To find the Remainder, find remainder of each of the terms when divided by 4 (871 * 873 * 7 * 439 * 881 * 883) and multiple them together:

871 mod 4 = 3
873 mod 4 = 1
7 mod 4 = 3
439 mod 4 = 3
881 mod 4 = 1
883 mod 4 = 3

(3 * 1 * 3 * 3 *1 * 3) = 81 mod 4 = 1

So N = 4M + 1

So S/250 = 4M + 1
S = 1000M + 250

Therefore, it must end in 250 and sum will be 7.
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Re: Find the sum of last three digits of S, where S= 871*873*875*878*881*8 [#permalink]
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Re: Find the sum of last three digits of S, where S= 871*873*875*878*881*8 [#permalink]
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