SAT: How to Solve IMPOSSIBLE Math Problems!

SAT, Study and Triumph: How to Solve IMPOSSIBLE Math Problems! Have you ever seen a math question that just seemed like it was too impossible to solve? Whether it was during a practice test, or during some assigned homework, you saw that problem and just thought, “Wait…what??” With this first webisode of “Study and Triumph”, Cambrian will help show you how those “impossible” math problems might not be so difficult after all, so you can walk in on test day confident that you’ll triumph in the Math section!

Cambrian Thomas-Adams is a 99th percentile SAT instructor for Veritas Prep. He recently graduated cum laude from Yale University with distinction as a Theater Studies major. He is happy to be able to continue the long tradition of teachers in his family and help students tackle the SAT!

Take advantage of Veritas Prep’s free SAT resources at

If you’d like to learn more about Veritas Prep’s SAT prep program and find a friendly tutor like Cambrian, go to

14 thoughts on “SAT: How to Solve IMPOSSIBLE Math Problems!

  1. That's pretty good for a young kid reading a script. Now let's see what a 40+ year tutor has to say.

    It doesn't matter what the whole number is 7 raised to the first, second, third, fourth etc power is.

    What matters is the last digit. Save time. Only worry about the last digit.

    7^1 = 7

    7^2 = 9 (only the last digit, the units digit, matters.)

    7^3 = 3

    7^4 = 1

    7^ 5 = 7

  2. That's…wrong…I'm pretty sure. The answer is 1 because the sequence ends its loop on 7, making 9 remainder 1, 3 remainder 2, and finally 1 on remainder 3. This is why 7^1 is 7. In your case 1 would never be reached.

  3. your resources are wesome veritas!!! id like to know the answer to one big problem though:given the equation 2X^2-6X+9+2y^2+2Y+1=45, how do i find the centre of the circle?

  4. OK, so I have a question.  Why did you start with 7^1 and not 7^0?  If you start with 7^0, then the pattern would be 1, 7, 9, 3.  The answer to the math question is the third number in the pattern, which in this case is 9.

  5. Wait, Wait ,WAIT.
    That was a ridiculous way of solving a problem as simple as this.All you need to do, is look for the unit's place.Thus,
    Unit's digit of: 7^1=7
    Unit's digit of: 7^2=Unit's digit of:7^1*7=9
    Unit's digit of: 7^3=Unit's digit of:7^2*7=3
    Unit's digit of: 7^4=Unit's digit of:7^3*7=1
    Unit's digit of: 7^5=Unit's digit of:7^4*7=7
    Unit's digit of: 7^6=Unit's digit of:7^5*7=9
    and so on,…..
    Thus in this way the pattern cam be found out very easily.

  6. the threee terms in a sequence goes 1(1*2)+1(2*3)+1(3*4)+…..
    the nth term of this sequence is 1(n)(n+1) . the qn is whats the sum of first 50 terms of this sequence

  7. Can you take the SAT in sixth grade I really want to take it just so I can get a heads up on it.

  8. Great video! I really enjoy seeing different perspectives on how to approach these kind of problems.

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