1001 Algebra Problems.PDF
- d.Multiplying 5 times 5 yields 25. Then, mul-
tiplying this product by 5 results in 125. Thus,
5 5 5= 125.
- c.By the definition of an exponent, we have
35 = 3 3 3 3 3 = 243.
- b.First, the following are the multiples of 6
between 0 and 180:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78,
84, 90, 96, 102, 108, 114, 120, 126, 132, 138,
144, 150, 156, 162, 168, 174, 180
Of these, the following are also factors of 180:
6, 12, 18, 30, 36, 60, 90, 180
There are eight possibilities for the whole
number p.
- c.The only choice that is a product of prime
numbers equaling 90 is 2 3 3 5.
- c.The factors of 12 are 1, 2, 3, 4, 6, 12. Of
these, only 1 and 3 are not multiples of 2.
Thus, the set of positive factors of 12 that are
NOT multiples of 2 is {1,3}.
- e.The sum of 13 and 12 is 25, which is an odd
number. Each of the other operations produces
an even number: 208 = 160, 37 + 47 = 84,
7 12 = 84, 36 + 48 = 84
- c.By the definition of an exponent, 2^4 = 2 2
2 2 = 16.
- b.Applying the order of operations, we first
perform exponentiation, then subtract from
left to right to obtain 9 – 2^2 = 9 – 4 = 5.
- c.The only choice that is divisible by only 1
and itself is 11. Each of the other choices has
factors other than 1 and itself.
Set 2 (Page 3)
- c.Begin by simplifying the absolute value
quantity. Then, divide left to right:
- 25|4 – 9| = – 25|–5| = –255 = –5.
- a.Since there are an odd number of negative
signs, the product will be negative. Computing
this product yields –4–2–63 = –144
- c.Applying the order of operations, first sim-
plify the quantity enclosed in parentheses,
then square it, then multiply left to right, and
finally compute the resulting difference:
5– (–17 + 7)^2 3 = 5 – (–10)^2 3 = 5 – 100
3 = 5 – 300 = –295
- a.Applying the order of operations, first com-
pute the quantities enclosed in parentheses.
Then, compute the resulting difference:
(497) – (48(–4)) = (7) – (–12) =
7 + 12 = 19
- d.Note that substituting the values 1, 2, and 3
in for p in the equation y= 6p– 23 yields –17,
–11, and –5, respectively. However, substitut-
ing 4 in for presults in the positive number 1.
So, of the choices listed, the least value ofpfor
which yis positive is 4.
- b.Applying the order of operations, first com-
pute the quantities enclosed in parentheses.
Then, compute the difference from left to right:
–(53) +(12( – 4)) = –(15) + (–3) =
–15 – 3 = –18
- c.Applying the order of operations, compute
both exponentiated quantities first. Then,
multiply from left to right as products arise.
Finally, compute the resulting difference:
–2(–2)^2 – 2^2 = –2(4) – 4 = –8 – 4 = –12
ANSWERS & EXPLANATIONS–
