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Number - Factors and multiple

SOURCE:GMAT
Number of Problems: 25.
FOR PRINT ::: (Book)

Problem Num : 11
Type: Understanding
Topic:Number 

Adjustment# : 21
Difficulty: 1

Category Factors and multiple
Analysis Divisible by 1 through 7 ? Multiple of 1 through 7?
Solution/Hint The number should have 1 through 7 as a factor. so.. 2*3*2*5*7*p so.. What is the smallest number then?


Problem Num : 12
Type:
Topic:Number 
Theme:None
Adjustment# : 22
Difficulty: 1

Category Factors and multiple
Analysis Pick a number
Solution/Hint


Problem Num : 13
Type: Mathematical Writing
Topic:Number 
Theme:None
Adjustment# : 22
Difficulty: 600 level

Category Factors and multiple
Analysis d is divisible by 25 d= 25k
Solution/Hint (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..


Problem Num : 14
Type:
Topic:Number 
Theme:None
Adjustment# : 23
Difficulty: 1

Category Factors and multiple
Analysis
Solution/Hint


Problem Num : 15
Type:
Topic:Number 
Theme:None
Adjustment# : 24
Difficulty: 1

Category Factors and multiple
Analysis 12, 30, 21.. (1)and (2) are equivalent
Solution/Hint


Problem Num : 16
Type: Understanding
Topic:Number 
Theme:None
Adjustment# : 25
Difficulty: 2

Category Factors and multiple
Analysis Two consecutive number is divisible by 2 Three consecutive number is divisible by 3.
Solution/Hint


Problem Num : 17
Type:
Topic:Number 
Theme:None
Adjustment# : 26
Difficulty: 1

Category Factors and multiple
Analysis
Solution/Hint


Problem Num : 18
Type:
Topic:Number 
Theme:None
Adjustment# : 26
Difficulty: 1

Category Factors and multiple
Analysis We can express the three consecutive integers as n, n+1, n+2
Solution/Hint


Problem Num : 19
Type:
Topic:Number 
Theme:None
Adjustment# : 30
Difficulty: 1

Category Factors and multiple
Analysis
Solution/Hint


Problem Num : 20
Type: Understanding
Topic:Number 
Theme:None
Adjustment# : 31
Difficulty: 600 level

Category Factors and multiple
Analysis 147000=3*7*7*2^3*5^3

what's your target to find?

Solution/Hint 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?



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