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08 May 2011, 03:31
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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
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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08 May 2011, 08:22
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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?
properties-of-divisible-numbers-79896.html
m I missing something?
08 May 2011, 11:57
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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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
