SAT – number properties

Q: Find the largest number less than 4000 that is the square of an integer.

Answer: 3,969

Explanation:  The simplest, fastest way to do this is:

On a calculator, find the square root of 4000.  It is 63.245….  Round this down to 63.  Now, square 63.  It is 3,969.

This is the way you should be thinking for SAT problems.  If you decide instead to count down from 3,999, finding the square root of each number until you find a whole number, you will have to do this 31 times until you find the right one!  (And by the way, why 31?  Doesn’t 3999 – 3969 equal 30?  Was I wrong?  No!  This is another little SAT trick.  Always add 1 to include the first or last number, whichever one wasn’t counted in the original 30.)

 

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SAT – number properties

Q: The variables a, x, and b are positive integers. If ax = 7, xb = 28, and ab = 4, what is the value of a?

Answer: 1

Explanation:

a, x, and b are all positive integers, so the only possibilities for the variables are factors of 7, 28, and 4.

Since 7 is prime, a and x must be 7 and 1, but in which order? If x = 1, then b = 28, and then we can’t find an integer value for a from the last equation. So, x must instead be 7.

It follows that a = 1 and b = 4.

The answer is 1.

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SAT – number properties

Q: What is the sum of all of the integers from 1 to 100?

Answer: 5050

Explanation: Don’t add 1 + 2 + 3 + 4 + … all the way up to 100! While it will work, it is a giant waste of time. Look for a pattern of numbers to pair up:

 1 + 99 = 100

2 + 98 = 100

3 + 97 = 100

49 + 51 = 100

 That’s 49 one hundreds. 49×100 = 4900. Now, add the numbers we have not counted yet: 100 and 50.

4900 + 100 + 50 = 5050.

 My TI-84 Plus program ‘SUMATOB’ will find this sum, too.

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SAT – number properties

Q: How many positive integers less than 500 are multiples of 3, 7, and 11?

Answer:  2

Explanation:  3, 7, and 11 do not have any common factors except 1.  So, their least common multiple is 3 x 7 x 11 = 231.  The next two common multiples are 231 x 2 = 462 and 231 x 3 = 693.  693 is greater than 500.  There are only two that are less than 500.  They are 231 and 462. 

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SAT – Remainders

Q: When the positive integer b is divided by 5, the remainder is 1. When b is divided by 7, the remainder is also 1. If b is less than 100, what is the largest possible value for b?

Answer: 71

Explanation: b is an integer that is one more than a multiple of both 5 and 7.  The smallest such value is 35 +1 = 36.  But, we need the largest one less than 100.  The next common multiple of 5 and 7 is 70 (do 35 x 2), and the next one is 105 (35 x 3).  Use 70, and add 1.  71.

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SAT – Number Properties

Q: A red light flashes every 10 minutes, and a blue light flashes every 6 minutes. If both lights flash together at 6:50 pm, when is the next time they will both flash together?
 
Answer:  7:20 pm
Explanation:  The least common mutiple of 10 and 6 is 30.  The light will both flash together again 30 minutes after 6:50, which is 7:20.
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SAT – Remainders

Q: When the positive integer k is divided by 6, the remainder is 2. What is the remainder when k+2 is divided by 6?
 
Answer: 4
Explanation:  Choose a number for k that leaves a remainder of 2 when divided by 6.  For example, k = 8  (one group of 6, with 2 left, equals 8).  Now do k+2, or 10, divided by 6.  6 divides into 10 once, with 4 left.  The remainder is 4.  My TI-84 Plus program ‘REMAINDR’ will find this remainder for you.
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