Probability+Power+Divisibility rules

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)

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.

[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.

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.