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by sana.noor » Sat May 11, 2013 12:54 am
If the integers m and n are chosen at random from 1 to 100, then what is the probability that a number of the form 7^m + 7^n is divisible by 5?
(A) ½
(B) ¼
(C) 1/6
(D) 1/8
(E) 1/16
B
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by GMATGuruNY » Sat May 11, 2013 2:46 am
sana.noor wrote:If the integers m and n are chosen at random from 1 to 100, then what is the probability that a number of the form 7^m + 7^n is divisible by 5?
(A) ½
(B) ¼
(C) 1/6
(D) 1/8
(E) 1/16
B
For 7^m + 7^n to be a multiple of 5, its units digit must be 0 or 5.

When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.

The units digits repeat in a CYCLE OF 4.
For exponents 1 through 100, this cycle of 4 will repeat 25 TIMES.
The result:
For exponents 1 through 100, each of the 4 units digits -- 7, 9, 3 and 1 -- will appear the SAME NUMBER OF TIMES.

Question rephrased:
If the units digits of 7^m and 7^n have an EQUAL CHANCE of being 7, 9, 3, or 1, what is the probability that 7^m + 7^n is divisible by 5?

Total options for 7^m + 7^n:
The number of options for the units digit of 7^m = 4. (7, 9, 3, or 1.)
The number of options for the units digit of 7^n = 4. (7, 9, 3 or 1.)
To combine these options, we multiply:
Total options = 4*4 = 16.

Good options for 7^m + 7^n:
For 7^m + 7^n to be divisible by 5, the sum of the units digit of 7^m and the units digit of 7^n must be 0 or 5.
The following combinations are good:
7+3
9+1
3+7
1+9
Good options = 4.

Thus:
(good options)/(total options) = 4/16 = 1/4.

The correct answer is B.
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by vipulgoyal » Sun May 12, 2013 11:07 pm
Hi Mitch I believe its Nowhere mentioned the the value of 7^m or value of 7^n must be below 100
its m and n could be any no fron 1 to 100


If the integers m and n are chosen at random from 1 to 100, then what is the probability that a number of the form 7^m + 7^n is divisible by 5?
(A) ½
(B) ¼
(C) 1/6
(D) 1/8
(E) 1/16


In order to divide by 5 the unit digit of 7^m + 7^n must be 5 or 0

we need to select m and n from 100
so total no of ways = 100x100 = 10000

now

the unit digit of 7 repetes after cycle of 4 -- 7,9,3,1,7,9,3,1
in order to get unit digit 0 for 7^m + 7^n we could have

unit digit 7 + unit didit 3
7^(1,5,9---97) AND 7^(3,7,11,--99)OR

unit digit 3 + unit didit 7
7^(3,7,11,--99) AND 7^(1,5,9---97)OR

unit digit 9 + unit didit 1
7^(2,6,10--98) AND 7^(4,8,12--100)OR

unit digit 1 + unit didit 9
7^(4,8,12--100) AND 7^(2,6,10--98)

There is no way to get unit digit 5 with 7^m + 7^n ( m&n can be any no from 1 to 100)

Therefor Required probabtlity

25x25 + 25x25 + 25x25 + 25x25

Hence the probablity = require probablity/ Total probablity
= 2500/10000 = 1/4
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by vipulgoyal » Sun May 12, 2013 11:42 pm
Experts please shed some light
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by mkdureja » Mon May 13, 2013 2:49 am
Unit digit of 7^m are in cyclical order with cyclicity of 4.
Since m can vary from 1 to 100, and 100 is divisible by 4, so every last digit , ie 7, 9, 3, 1 all are distributed equally.
Therefore you can safely do the maths for only 4 numbers, probability wont change if you are multiplying both the (good options) and (total options) by the same number(25 in this case).
So, answer = 4/16 = 1/4 (as Mitch has explained)
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