War on Integers

N is a multiple of 5 only if the units digit is a 0 or 5. So, looking at the units digits only.

The patterns are:

For 1:
1
1
1
1

For 2
2
4
8
6

For 3
3
9
7
1

For 4
4
6
4
6

Together they look as follows:

1..2..3..4
1..4..9..6
1..8..7..4
1..6..1..6

And this pattern will continue

1+2+3+4 = 10 > Multiple of 5

1+4+9+6 = 20> Multiple of 5

1+8+7+4 = 20 > Multiple of 5

1+6+1+6 = 14 > Not Multiple of 5

Since this pattern continues, ¾ of the numbers will be multiples of 5. I will go with B, 75.

Is this correct?

Ingenious. I tried to find the pattern and deduced that because all three Pn's for n=1, n=2, and n=3 are divisible by 5, then all 100 integers will give a Pn that is divisible by 5. Boy am I in trouble... :roll:

dmateer25, where did you get this question? Do they really have such complex ones on the GMAT?

for the number to be a multiple of 5, the units digit has to be 5 or 0;

1^n will always yield 1 in unit's digit
2^n can yield 2, 4, 8 , 6 and then this pattern will repeat
3^n can yield 3, 9, 7, 1 and then repeats
4^n can yield 4, 6 and then repeats

so for n =1,2,3 ... we will have the below pattern in unit's digit

1 1 1 1
2 4 8 6
3 9 7 1
4 6 4 6
--------------
10 20 20 14

for every 4 nos, 3 will yield a multiple of 5 therefore 100 will have 75 such integers for which given eqn is a multiple of 5

dtweah wrote:Let Pn = 1^n + 2^n + 3^n + 4^n. Find the number of integers n for which
1 ≤ n ≤ 100 and Pn is a multiple of 5.
(a) 68
(b) 75
(c) 86
(d) 98
(e) 100

I uploaded the solution which must be studied b/c it has some important tricks or properties. I did not approach it that way.

The first thing to notice about Pn is that its sum is always even: two odd + two even numbers is even.( Look out for these kinds of generalizations) So the only even number divisible by five in the problem is one that ends in 0. How many are there? I tested the first 3 and observed that all ended in 0. But the number "4" failed, But 5 succeeded so I thought there must be a pattern in which every 3 number ends in 0 and the fourth one doesn't.

So 1 2 3 4 5 6 7 8 9 10 11 12.... 100

4 8 12 16 20 24....100 are failing

Notice the numbers that are failing. They are just your 4 times table.

Since 4 x 25 =100, I reasoned there must be 25 such numbers that cannot produce a zero at the end.

So B.

@dtweah, Can u check the attached doc??

kyabe wrote:@dtweah, Can u check the attached doc??

Sorry Kyabe. This should do now.