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# Remainder, you are awesome ....

Remainders are wonderful and you can do wonders with remainders in solving some otherwise looking dreadful questions of Real Numbers.

See this example. Suppose, you divide 15 by 7, you have remainder 1, and you can always write it in terms of Euclid Division Algorithm, as:

15 = (2x7) + 1. Though the remainder is always a positive integer, you can always think it writing like this too: 15 = (3x7) + (-6).

And this can be written as: 15 == -6 mod 7. This idea is wonderful in solving many questions. See a wonderful use of this idea.

Qu: Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

Sol. Let the number be P. Then, as given:

P = 4 mod 5, i.e. P = -1 mod 5

P = 5 mod 6, i.e. P = -1 mod 6

P = 6 mod 7, i.e. P = -1 mod 7

P = 7 mod 8, i.e. P = -1 mod 8

P = 8 mod 9, i.e. P = -1 mod 9

Means, the number leaves a remainder -1 in each case when divided by 5, 6, 7, 8, 9.

Hence, the number is:

LCM (5, 6, 7, 8, 9) + (-1) = LCM {LCM (5,6,7), LCM (8,9)} - 1

Since 5,6,7 are co-prime; and 8, 9 are co-prime, we have:

LCM {LCM (5,6,7), LCM (8,9)} - 1 = LCM {210, 72} - 1 = {6x35x12} -1 = 2520 – 1 = 2519 {Look at the way the LCM has been worked out}

The required number is: 2519

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