Examples: Monday, today, last week, Mar 26, 3/26/04
We've updated the Topics page of our website to better organize and share our expert content. Read more about it here.

1266 replies24 threadsRegistered User Senior Member
I thought I would start a thread of math problems that students have asked me about while preparing for their SAT. After I post each question I'll give some time for you to post your attempted solutions before providing my own. I'll try to provide the Difficulty Level and Topic for each one. Here is the first one:

Level 3 Number Theory

1. If an integer n is divisible by 15, and 25, what is the next larger integer divisible by these numbers?

(A) n + 15
(B) n + 50
(C) n + 75
(D) n + 125
(E) n + 150
636 replies

## Replies to: SAT Math Problems Thread

• 112 replies21 threadsRegistered User Junior Member
• 1266 replies24 threadsRegistered User Senior Member
That's correct. Can you tell us how you did it?
• 105 replies19 threadsRegistered User Junior Member
edited April 2014
I have a fast yet somewhat ridiculous solution here. I just find a common number that is divisible by 15 and 25 see if any fits the letter from A to E
edited April 2014
• 81 replies15 threadsRegistered User Junior Member
edited April 2014
Since it is a level 3 problem, one should expect that it wont be that easy and any answer that attracts you instantly is probably a wrong answer.For example, n+50 seems attempting for me since it is the next larger integer that is a multiple of 5.

How did I do it?

GCF x LCM= 15 x 25

GCF is for sure equal to 5.

(25 x 15)/ 5 = 75 ,so the answer is C.

edited April 2014
• 1266 replies24 threadsRegistered User Senior Member
edited April 2014
@Sparkkid1234 Not a ridiculous solution at all. You are simply taking the lcm of 15 and 25 which is 75 - a nice quick and elegant solution.

@meumeu Your solution is similar. You didn't have to do quite so much work. Computing the lcm directly is just a bit quicker than computing both the product and the lcm.
edited April 2014
• 1266 replies24 threadsRegistered User Senior Member
This problem can also be done by picking numbers. If you're comfortable finding the lcm, then you can use 75 for n. If not, the product 375 will work as well.
• 1266 replies24 threadsRegistered User Senior Member
Level 3 Number Theory

2. Dana has pennies, nickels and dimes in her pocket. The number of dimes she has is three times the number of nickels, and the number of nickels she has is 2 more than the number of pennies. Which of the following could be the total number coins in Dana’s pocket?

(A) 14
(B) 15
(C) 16
(D) 17
(E) 18
• 2355 replies173 threadsRegistered User Senior Member
d = 3n, n = 2p so the total number of coins has to be a multiple of 9 (the minimum is 1 penny, two nickels and 6 dimes). The only multiple of 9 in the answer set is e) 18.
• 4728 replies19 threadsRegistered User Senior Member
^ The problem says the number of nickels is 2 more than the number of pennies (not 2x more).

Letting p be the number of pennies, then the number of coins is p + (p+2) + 3(p+2) = 5p + 8. The number of coins must be 3 (mod 5) (i.e. 3 more than a multiple of 5) so (E) 18 is the answer.
• 2355 replies173 threadsRegistered User Senior Member
^ Thanks. I agree with you as far as 5p + 8. At that point, wouldn't it make sense just to plug in numbers for p. Starting with p=1 or total coins = 13. Since that is not an answer option move on to p=2 and total coins = 18. 18 is answer E) so you're done.
• 81 replies15 threadsRegistered User Junior Member
^^ That's genius.However, the answers may come as variables not numbers.
When you practice any question, it is generally better to know how to do it in 2 ways.
• 112 replies21 threadsRegistered User Junior Member
Yeah, I always just plug in numbers that meet whatever is stipulated in the question.
• 2355 replies173 threadsRegistered User Senior Member
^Plugging in is a great strategy, but it doesn't always make sense. Some problems don't lend themselves to plugging in and other problems can be solved easier using other strategies.
• 58 replies8 threadsRegistered User Junior Member
edited April 2014
Here's another one:

3 men and 3 women stand in 1 line. Two or more of the same gender cannot stand next to each other. How many different arrangements are possible?
edited April 2014
• 112 replies21 threadsRegistered User Junior Member
_M_M_M_
4! x 3! = 144 arangements
• 4728 replies19 threadsRegistered User Senior Member
@CandyPants16 there are three men and three women.
• 1059 replies20 threadsRegistered User Senior Member
It's a nice question. You can solve it with the counting principle. You can do it in two separate cases but you don't have to. Just go spot by spot asking "Now how many choices are there?"
• 1266 replies24 threadsRegistered User Senior Member
I think Candypants solution has a small error. The picture makes it look like there are 4 configurations, but there are only 2: _M_M_M and M_M_M_

So there are 3! ways to arrange the males, 3! ways to arrange the females, and 2 possible ways to configure the males and females. By the counting principle, we get 2*3!*3! = 72

(As an example of the error, the configuration _M_MM_ has two males next to each other)

Pckeller's method gives 6*3*2*2*1*1 = 72 as well.