Problem Solving

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

sanju09 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

_________
Hey sanju09,
The same problem was posted by me a couple of days back.Please do not re-post the same problem again and again.It would be definitely appreciated in the interest of the forum.

harsh.champ wrote:

sanju09 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

_________
Hey sanju09,
The same problem was posted by me a couple of days back.Please do not re-post the same problem again and again.It would be definitely appreciated in the interest of the forum.

Alright!! And what did you answer? Give us the link at least.

sanju09 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

If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3 which is not in options

sanju09 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

ok sanju and harsh I will give u the answer of this question, its quite easy:

(7^k) /5 gives remainder = 2 for integers n of the form 4p+1;
= 4 for integers n of the form 4p+2;
= 3 for integers n of the form 4p+3;
= 1 for integers n of the form 4p;
now from numbers 1 - 100 (since 100 is divisble by 4) so we have 1/4th of each of 4p,4p+1, 4p+2,4p+3.

Now for the 7^m+ 7^n to be divisble by 5 , the sum of the remainders must be divisble by 5.

so we check for probability of the sum of any two of 2,4,3,1 being divisible by 5 .
so its probabilty = 4/16 = 1/4
(2+3,4+1,1+4, 3+2) out of 4^2=16 numbers that could be formed from the coupling of 2,4,3,1.

Thus overall probabaility of 7^m + 7^n to be divisible by 5 = 1/4[/quote]

ajith wrote:

sanju09 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

If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3 which is not in options

I made a mistake in the highlighted portion
One can select the same no twice hence the no of ways =8
and thus the probability becomes 2/8 =1/4

Thanks to Shashank

ajith wrote:
sanju09 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
If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3 which is not in options
I made a mistake in the highlighted portion
One can select the same no twice hence the no of ways =8
and thus the probability becomes 2/8 =1/4

Thanks to Shashank

yeah ajith I was about to point your mistake..."one can select two numbers from it in = 6 ways "
That is good you pointed it out yourself..

sanju09 wrote:

harsh.champ wrote:

sanju09 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

_________
Hey sanju09,
The same problem was posted by me a couple of days back.Please do not re-post the same problem again and again.It would be definitely appreciated in the interest of the forum.

Alright!! And what did you answer? Give us the link at least.

__________________________
Hey sanju09,
Here's the link of the question I had posted before under the subject:Probability+Power+Divisibility rules.
This thread was just 2 threads above you.
Also,ajith and shashank.ism,maybe you would like to look at the answer posted by Ian Stewart.

harsh.champ wrote: _____________
Hey sanju09,
Here's the link of the question I had posted before under the subject:Probability+Power+Divisibility rules.
This thread was just 2 threads above you.
Also,ajith and shashank.ism,maybe you would like to look at the answer posted by Ian Stewart.

Harsh, Ian answered a totally different question ; also, Shashank has given the link to this thread in that thread.

Ajith how to find the number oif ways 8?

stufigol wrote:Ajith how to find the number oif ways 8?

It is not actually 8 ways - My solution above is flawed.

1,3,7,9 can be the last digits of the number with equal probability

One can choose 2 numbers from this with replacement = 4*4 ways =16 ways
out of which if you select pairs (7,3) (3,7) (1,9) (9,1) the resulting number would be divisible by 5

Probability of that happening = 4/16 =1/4

I suggest you take a look at shashank's solution

ajith wrote:

ajith wrote:

sanju09 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

If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3 which is not in options

I made a mistake in the highlighted portion
One can select the same no twice hence the no of ways =8
and thus the probability becomes 2/8 =1/4

Thanks to Shashank

I found your explanation very clear so I understood your approach but could you please explain why there are 8 different ways in selecting the 2 unit digit numbers from the set 7,3,9,1. i actually find out 6 ways to selct these numbers...

hey, can I do it this way:

to select m: 100C1 ways, any number can be chosen

to select n: 25C1 ways;
m=1 units place 7
m=2 units place 9
m=3 units place 3
m=4 units place 1
--------------------------
m=5 units place 7
m=6 units place 9
m=7 units place 3
m=8 units place 1
-------------------------
..
..
m=99 units place 3
m=100 units place 1
----------------------------

we have 25 such groups, thus we have 25 numbers that will give unit digit as 7, 25 numbers that will give unit digit as 9, 25 numbers that will give unit digit as 3, 25 numbers that will give unit digit as 1

so each value of m will have corresponding 25 numbers to choose from, hence n can be chosen in 25 ways

total ways 100C1 100C1

answer (100 x 25) / (100 x 100) = 1/4

Sorry don't know how to explain it in a better way....

will it work if I approach it with the method below
7^m+ 7^n = 7^m(1+7^a) where m= n+a
(1+7^a) will have 8, 0(9+1), 4 and 2 as the last digit cos' 7^a will have 7,9,3 and 1 as the last digit

of 2(a=0), 8(a=1), 0(a=2) and 4(a=3) only when the unit digit is 0 will the eq be divisible by 5

so it is 1/4

ajith wrote:

sanju09 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

If a number is divisible by 5 - it should either end at 0 or 5

the possible values last digit of 7^m is = 7,9,3,1

one can select two numbers from it in = 6 ways

2 favourable conditions are = 7,3 and 9,1

Probability =1/3 which is not in options

You almost solved it dude:

7^m can have last digit as 7,9,3,1
and
7^n can have last digit as 7,9,3,1

favorable values are (7,3), (9,1), (3,7) and (1,9) out of 16 possible pairs.
Thus, required probability =

4/16 = 1/4

The catch here is that since all the numbers 7,9,3,1 will occur exactly equal number of times, thus my above explanation will hold. The value 100 is just to serve this purpose.