hard sat math problems?

Test Preparation SAT Preparation
1

1.What is the units digit of 9^999+10^1000+11^1001

2.For all positive integers n, let #n be the remainder when n is divided by 9. Which of the following is equal to #123?
A. #234
B. #345
C. #567
D. #678
E. #789

3.For how many integers n between 300 and 400 is n^(2/3) also an integer?

  1. What is the least possible three-digit integer with three different digits that are all multiples of 3?

I put the 369 but its wrong.

2
  1. the remainder is 6 when 123/9 789/9 the remainder is 6
    3)answer is 1 because the only number that has a cubed value between 300 and 400 is 7
3

are you sure 369 is wrong?

4
  1. 9^1 has a units digit of 9 9^2 has a units digit of 1 9^3 has a units digit of 9 ... 9^999 has a units digit of 9

10^1000 has a units digit of 0, and 11^1001 has a units digit of 1.

Units digit of 9^999 + 10^1000 + 11^1001 = 0

More formally, 9 ≡ -1 (mod 10), so 9^(999) ≡ (-1)^999 ≡ -1 (mod 10).

  1. Note that a positive base 10 integer N is congruent to the sum of its digits mod 9, i.e. N and the sum of the digits of N leave the same remainder when divided by 9. Since 1+2+3 = 6, we are looking for an integer whose sum of digits leaves a remainder of 6 when divided by 9. Choice E works, as 7+8+9 = 24 ≡ 6 (mod 9).
  2. If n^(2/3) = k (for some integer k), then n^2 = k^3, which means that n must be a perfect cube (do you see why?). Only 7^3 lies between 300 and 400, so 1 integer.
  3. 0 is also a multiple of 3 --> 306.
5

yeah the answer is 309 apparently?
edit: 306

6

@catepillargiraf 306

7

would you rate these problems as level 5 ?

8

why would you divide 123 by 9 if it says 123 is the remainder?

9

@catepillargiraf 1. Most likely. Unless the student knows modular arithmetic or that the units digits of powers cycle, he would have to try finding the units digits of small powers first.

  1. I don't think so, since it's really easy to just compute #123 = 6 and check the answer choices.
  2. Maybe. You'd have to establish that n is a perfect cube, and that non-cube integers don't work.
  3. Probably not, although I can see how one can miss 306. I'd say maybe a tricky level 3 or 4.
10

Bottle A contains 100 cubic centimeters of a solution that is 60% water and 40% acid. Bottle B contains 100 cubic centimeters of a solution that is 80% water and 20% acid. Twenty cubic centimeters are poured from bottle A into Bottle B, and then the contents of bottle B are then thoroughly mixed. Then 30 cubic centimeters are poured from bottle B into Bottle A. What fraction of the solution in bottle A is acid?

this seems to be a multi-step problem… but i don’t know where to start it off

11

the answer is 39/110 my explanation may be confusing.bottle a ratio is 6:4 (water:acid)if you pour 20% of bottle a to b you give bottle b 12%water and 8%acid which would make bottle b 120 cubic centimeters and the percentage 23/30 water and 7/30 acid and then you pour back 30 centimeters so bottle a is now 110 with 23 extra water and 7 extra acid
so bottle a has 110 with 71 water (48+23) and 39 acid(32+7) 39/110

12

how’d you get 23/30 and 7/30 and whered you get the initial value of 48 water and 32 acid? sorry. im lost

13

I got 48 and 32 from subtracting the original value with the value that gets poured in bottle b 60-12 and 40-8
bottle b originally had 100 cubic cm with 80 water and 20 acid but when it got 20 extra it now had 120 cubic cm with 92 water and 28 acid 92/120 and 28/120 is 23/30 and 7/30 respectively