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The lucky number trick: Remainder when divided by 9

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Status: 700 (q47,v40); AWA 6.0
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The lucky number trick: Remainder when divided by 9 [#permalink] New post 08 May 2011, 02:31
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Everyone feels charmed about their lucky number. They take all pains to add up the digits in their date of birth and get to their licky number. A tiny mathematical trick is quite useful here. In the context of GMAT, I use it extensively to attack remainder problems when large numbers are involved. Say when I have to find out if 3 is one of the prime factors, I use this trick.

This post is deliberately being left out of the Remainders post which is more elaborate and covers more prime numbers than the explanation below entails.

Whenever I have to find the remainder divided by 9, I merely add all the digits and take the sum. I subject the sum to the same trick again and this process goes on repeatedly until I have a single digit. That single digit is the remainder when the number is divided by 9. For example, consider the following number:

723965402

The first sum would be: 7 + 2 + 3 + ... + 2 = 38
Repeating on 38, I get a 11 and repeat on 11, the final sum will be 2 which means that the number leaves a remainder of 2 when divided by 9. This also means that the number is not divisible by 3 and leaves a remainder of 2.

What does this mean to GMAT math questions? If you want to quickly find out whether 3 is a prime factor, use this trick and if 3 or 6 is the lucky number, then bingo. 3 is a prime factor involved.

Deciding about 2 and 5 needs no explanations. I am not sensing that the GMAT will test you on higher primes anyway.

NB: As a further shorter cut, do not add 9 if it appears somewhere in the digits. And even drop those numbers that add up to 9. You'd not need a calculator. So in the above number, the result is easy to see even as you glance at the number.

This also has a mathematical proof. So feel safe and exploit the trick

Regards
Rahul
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Rahul

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Re: The lucky number trick: Remainder when divided by 9 [#permalink] New post 08 May 2011, 07:22
if the sum is 38..that itself means it is not divisible by 3...as per the rule the sum of digits should be divisible by 3 then only the number is divisible by 3.

properties-of-divisible-numbers-79896.html

m I missing something?
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Re: The lucky number trick: Remainder when divided by 9 [#permalink] New post 08 May 2011, 10:57
I am saying decompose further as long as you would not have a single digit sum. 38 goes further down as 11 and that boils down to 2 which is not divisible by 9. Hence 38 is not divisible by 9. for simple 2 digit numbers, this trick is a good-for-nothing. The approach is specifically useful with very large numbers.

If however the final single digit sum is 3 or 6 (of course or 9 too), then we could say that 3 is a definite factor to that number.

[The mathematical theorem underlying the trick is related to congruences and may be a little too irrelevant here. Basically it says that if a number n leaves a remainder r when divided by a dividend d, then for any power of n the remainder will be the same as what you would have when r^n is divided by d].

When 10 is divided by 9, it leaves a remainder 1. When 9 divides 10^n, it would leave a remainder 1^n which is 1. So if a multi-digit number were to be written as x0 + x1*10^1 + x2*10^2 +.. + xi*10^i +.. then when divided by 9, the remainder is x1 + x2 + x3...+xi+.... Repeatedly applying this on the sum, one would arrive at the single digit which would be called the lucky number.

Regards
Rahul
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Regards
Rahul

Re: The lucky number trick: Remainder when divided by 9   [#permalink] 08 May 2011, 10:57
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The lucky number trick: Remainder when divided by 9

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