Hard GRE Math Arithmetic Problems with Solutions

Hard GRE Math Arithmetic Problems with Solutions

Yesterday I posted some really hard GRE Arithmetic problems, and today I would like to give you solutions to those problems. Please feel free to post your own solutions or attempted solutions in the comments below as well. If you still want further explanation after reading the below solutions please do not hesitate to ask.

Level 5 GRE Arithmetic Problems with Solutions

If n is a positive integer such that the units (ones) digit of n²+ 4n is 7 and the units digit of n is not 7, what is the units digit of n + 3?

Solution: By plugging in values for n, we find that for n = 9,

n² + 4n = 92 + 4 ⋅ 9 = 81 + 36 = 117.

So n = 9 works, and n + 3 = 9 + 3 = 12. So the units digit of n + 3 is 2.

Advanced solution showing the independence of n:

n² + 4n = n(n + 4).

So we are looking at positive integers 4 units apart whose product ends in 7. Since 7 is odd, n must be odd. So n must end in 1, 3, 5, or 9. Note that we skip n = 7 since the problem forbids us from using it.

If n ends in 1, then n + 4 ends in 5, and n(n + 4) ends in 5.
If n ends in 3, then n + 4 ends in 7, and n(n + 4) ends in 1.
If n ends in 5, then n + 4 ends in 9, and n(n + 4) ends in 5.
If n ends in 9, then n + 4 ends in 13, and n(n + 4) ends in 7.

So n ends in a 9, and n + 3 ends in a 2.

The sum of the positive odd integers less than 200 is subtracted from the sum of the positive even integers less than or equal to 200. What is the resulting difference?

We write out each sum formally, line them up, and subtract term by term.

2 + 4 + 6 + … + 200
1 + 3 + 5 + … + 199
1 + 1 + 1 + … + 1

Now notice that we’re adding 1 to itself 100 times. So the answer is 100.

Note: It is easiest to see that we are adding 100 ones by looking at the sum of the positive even integers less than or equal to 200. There are 200/2 = 100 terms in this sum.

Quick solution: Once you get a little practice with this type of problem you can simply compute 100•1 = 100.

Solution using the sum feature on your graphing calculator: (This solution cannot be used on the GRE, but it could for other standardized tests where a graphing calculator is allowed) Press the 2nd button followed by the List button (same as Stat button).
Go to Math and select 5: sum( or press 5.
Press 2nd followed by List again.
Go to Ops and select 5: seq( or press 5.
Enter x, x, 1, 199, 2)).
The display should look like this: sum(seq(x, x, 1, 199, 2)).
Press Enter and you should get the answer 10,000.
Next enter sum(seq(x, x, 2, 200, 2)) and you should get the answer 10,100. Finally, 10,100 – 10,000 = 100.

Note: In the expression sum(seq(x, x, 2, 200, 2)) the last 2 indicates the step size. Here we are adding every other number.

The positive number k is the product of four different positive prime numbers. If the sum of these four prime numbers is a prime number greater than 20, what is the least possible value for k?

Solution: Let’s begin listing sequences of 4 prime numbers, and checking if their sum is also prime, beginning with the smallest primes.

2, 3, 5, 7 Sum = 17 too small
2, 3, 5, 11 Sum = 21 not prime
2, 3, 7, 11 Sum = 23 prime
2, 3, 5, 13 Sum = 23 prime

Now, (2)(3)(7)(11) = 462, and (2)(3)(5)(13) = 390. Since 390 is smaller, k = 390.