BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability+Power+Divisibility rules

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

Probability+Power+Divisibility rules

by harsh.champ » Thu Feb 04, 2010 5:53 am
If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7^m + 7^n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Also,
If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these
__________________
Problem solving algorithm needed?Seeking advice.
Last edited by harsh.champ on Sat Feb 06, 2010 12:50 am, edited 3 times in total.
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Thu Feb 04, 2010 11:23 am
harsh.champ wrote:If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Problem solving algorithm needed?Seeking advice.
7m + 7n = 7(m+ n). If this is divisible by 5, then m+n must be divisible by 5, since 7 is not. So the question is asking: if m and n are chosen randomly from between 1 and 100, what is the probability that m+n is divisible by 5.

Now, I assume the question means "If m and n are chosen randomly from between 1 and 100 inclusive"; the question should make clear whether 1 and 100 are permissible values of m and n. If the range is inclusive, then the answer will be 1/5. There are a few ways to see this; we could consider units digits, since we're dealing with division by 5. No matter what I choose for m, there will be two possible units digits for n that will make m+n a multiple of 5. So for example, if m=23, then the units digit of n must either be 2 or 7 in order that m+n ends in 5 or 0. Since there are ten possible units digits for n in total, and two that make m+n a multiple of five, there is a 2/10 = 1/5 chance I choose a value of n that makes m+n divisible by 5.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

by harsh.champ » Thu Feb 04, 2010 1:05 pm
Ian Stewart wrote:
harsh.champ wrote:If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Problem solving algorithm needed?Seeking advice.
7m + 7n = 7(m+ n). If this is divisible by 5, then m+n must be divisible by 5, since 7 is not. So the question is asking: if m and n are chosen randomly from between 1 and 100, what is the probability that m+n is divisible by 5.

Now, I assume the question means "If m and n are chosen randomly from between 1 and 100 inclusive"; the question should make clear whether 1 and 100 are permissible values of m and n. If the range is inclusive, then the answer will be 1/5. There are a few ways to see this; we could consider units digits, since we're dealing with division by 5. No matter what I choose for m, there will be two possible units digits for n that will make m+n a multiple of 5. So for example, if m=23, then the units digit of n must either be 2 or 7 in order that m+n ends in 5 or 0. Since there are ten possible units digits for n in total, and two that make m+n a multiple of five, there is a 2/10 = 1/5 chance I choose a value of n that makes m+n divisible by 5.
__________________
Thanks.
I really liked the approach used by you.Actually,I have developed some fear whenever I see any divisibility question.
Just have to dispel my doubts.

So,I guess then the answer would be 5(none of the above as 1/5 is not mentioned)
Top
User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Fri Feb 05, 2010 7:45 am
harsh.champ wrote:
If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Problem solving algorithm needed?Seeking advice.

Harsh you have posted a good question but I think this question is not exactly this way. As to find 7m+7n to be divisible by 5 is very easy and you only have to check out divisibility of m+n by 5 except for the confusion that 1 and 100 are inclusive or not as pointed out by Ian stewart.

But I think the exact question would be bit tricky and it must be 7^m+7^n instead of 7m+7n. So please check your question again. In any case if the question is 7^m+7^n. You can find the answer in the following thread where I have posted its solution.

https://www.beatthegmat.com/7-m-7-n-is-d ... tml#221039

As for your fear against problem of division just solve a few question of such type and check the steps carefully. You would easily overcome it by practicing a few questions of such type.
Top
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Thu Sep 25, 2008 6:10 pm
Thanked: 2 times

by chintudave » Fri Feb 05, 2010 11:50 pm
[quote="Ian Stewart"][quote="harsh.champ"]If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Problem solving algorithm needed?Seeking advice.[/quote]

7m + 7n = 7(m+ n). If this is divisible by 5, then m+n must be divisible by 5, since 7 is not. So the question is asking: if m and n are chosen randomly from between 1 and 100, what is the probability that m+n is divisible by 5.

Now, I assume the question means "If m and n are chosen randomly from between 1 and 100 [b]inclusive[/b]"; the question should make clear whether 1 and 100 are permissible values of m and n. If the range is inclusive, then the answer will be 1/5. There are a few ways to see this; we could consider units digits, since we're dealing with division by 5. No matter what I choose for m, there will be two possible units digits for n that will make m+n a multiple of 5. So for example, if m=23, then the units digit of n must either be 2 or 7 in order that m+n ends in 5 or 0. Since there are ten possible units digits for n in total, and two that make m+n a multiple of five, there is a 2/10 = 1/5 chance I choose a value of n that makes m+n divisible by 5.[/quote]


Ian, I did not understand the logic behind reducing 7^m + 7^n to 7(m+n). Can you please ellaborate. I still think 1/4 is the right answer.
Top
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

by harsh.champ » Sat Feb 06, 2010 12:50 am
chintudave wrote:
Ian Stewart wrote:
harsh.champ wrote:If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7m + 7n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

____________________
Problem solving algorithm needed?Seeking advice.
7m + 7n = 7(m+ n). If this is divisible by 5, then m+n must be divisible by 5, since 7 is not. So the question is asking: if m and n are chosen randomly from between 1 and 100, what is the probability that m+n is divisible by 5.

Now, I assume the question means "If m and n are chosen randomly from between 1 and 100 inclusive"; the question should make clear whether 1 and 100 are permissible values of m and n. If the range is inclusive, then the answer will be 1/5. There are a few ways to see this; we could consider units digits, since we're dealing with division by 5. No matter what I choose for m, there will be two possible units digits for n that will make m+n a multiple of 5. So for example, if m=23, then the units digit of n must either be 2 or 7 in order that m+n ends in 5 or 0. Since there are ten possible units digits for n in total, and two that make m+n a multiple of five, there is a 2/10 = 1/5 chance I choose a value of n that makes m+n divisible by 5.

Ian, I did not understand the logic behind reducing 7^m + 7^n to 7(m+n). Can you please ellaborate. I still think 1/4 is the right answer.
_______________________
Hey chintudave,
Actually Ian did not reduce 7^m + 7^n to 7(m+n).By mistake while editing the post for the 2nd time,I left out the "^" in the 2nd question type.[My apologies to all]Two type of questions were posted.
1)Considering 7m + 7n
2)Considering 7^m + 7^n

Maybe Ian answered the 1st one. :)
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Sat Feb 06, 2010 1:58 am
chintudave wrote: Ian, I did not understand the logic behind reducing 7^m + 7^n to 7(m+n). Can you please ellaborate. I still think 1/4 is the right answer.
Yes, as harsh.champ points out, he has edited his original post since I first replied. As you will see in the text I quoted in my post above, he originally had written 7m + 7n; written that way, the answer is 1/5. If instead the question says 7^m + 7^n, as it does now in the edited post above, the answer will be different. To be completely clear, you certainly could never replace 7^m + 7^n with 7(m+n), or even with 7^(m+n); those are very different things.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Post new topic Post Reply
• Page 1 of 1