Topic : Number-Factors and multiple
1.
2.
3.
from: category_eng

1.None

Ans: e

2.Modification
2k+3m=t
1. Modify the given information.
2k+3m=t

(1) If k i s the multiple of 3, then k=3p
so.. t= 2*3p+3m =3*(2p+m)

(2) m=3q
then t=2k+3*3m. In this case, we can't have any idea about factor of t.

Ans: A

3.

Ans: C

1.


2.
There are several even number and multiple of 3. Why is the 36 a multiple of 3?

3.
LCM problem

4.


5.
81=3*3*3*3

6.


7.
How many 3s in 30!? at least 10.. In addition to, think about multiple of 9 and multiple of 27

8.
7150=2*5*5*(11*13)

9.
Divisible by 1 through 7 ? Multiple of 1 through 7?

10.
Pick a number

11.
d is divisible by 25 d= 25k

12.


13.
12, 30, 21.. (1)and (2) are equivalent

14.
Two consecutive number is divisible by 2 Three consecutive number is divisible by 3.

15.


16.
We can express the three consecutive integers as n, n+1, n+2

17.
147000=3*7*7*2^3*5^3

what's your target to find?


18.
96=3*2^5

19.
Factorize 3150

20.
(10x+y)+(10y+x)=11(x+y)

21.
15x+29y=440. 29y=5(88-3x) y should be multiple of 5. but less than 20.

22.


1.
2. E
Understanding
There are several even number and multiple of 3. Why is the 36 a multiple of 3?
(1) Even numbers: 36, 38, 42
(2) Multiple of 3 : 33, 36,39,42

Even though we take both, we can't answer for the question.
3. B
Understanding
LCM problem
x: a number of students in the class.
It's multiple of 8 and 12.

x=8p=12q.
2*2*2p 2*2*3q
4. D
Understanding
5. e
Understanding
81=3*3*3*3
x should be a factor of 81.
(1) x can be 3*3, 3*3*3, 3*3*3*3 to make it integer.
(2) X should be a multiple of 3
6. C
7. C
How many 3s in 30!? at least 10.. In addition to, think about multiple of 9 and multiple of 27
8. D
7150=2*5*5*(11*13)
9. A
Understanding
Divisible by 1 through 7 ? Multiple of 1 through 7?
The number should have 1 through 7 as a factor.
so.. 2*3*2*5*7*p
so.. What is the smallest number then?

10. A
Pick a number
11. E
Mathematical Writing
d is divisible by 25 d= 25k
(1) d=25p =5*5p
(2) d= 40q=2*2*2*5q

If we take both conditions, d is a multiple of 2*2*2*5*5.
so..
12. B
13. E
12, 30, 21.. (1)and (2) are equivalent
14. E
Understanding
Two consecutive number is divisible by 2 Three consecutive number is divisible by 3.
15. A
16. E
We can express the three consecutive integers as n, n+1, n+2
17. D
Understanding
147000=3*7*7*2^3*5^3

what's your target to find?


This is the same question of finding #of factor 7.
147000=3*7*7*2^3*5^3
so.. There are two 7's.




Discussion) How about 147000000?

18. A
96=3*2^5
19. E
Factorize 3150
20. A
(10x+y)+(10y+x)=11(x+y)
21. A
15x+29y=440. 29y=5(88-3x) y should be multiple of 5. but less than 20.
22. B
Understanding
n=m*k?
(1)3n=m*p
(2)13n=m*q, m is less than 13 so that it can't be a multiple of 13, which means that q is a multiple of 13. Therefore, n is a multiple of m.