Number properties

[Bright]

1) For how many positive integers x is 130/x an integer?

 

a) 8 b) 7 c) 6 d) 5 e) 3

 

 

2) If both the products and sum of four integers are even, which of the following could be the number of even integers in the group?

 

I. 0 II. 2 III. 4

 

a) I only b) II only c) III only d) II and III e) I, II, III

 

 

3) A wire is cut into 3 equal parts. The resulting segments are then cut into 4,6 and 8 equal parts respectively. If each of the resulting segments has an integer length, what is minimum length of the wire?

 

a) 24 b) 36 c) 48 d) 54 e) 72

 

OA to follow.....

[Ankost]

1) D

130/x=13*5*2/x

130 can be divided by 1,2,5,13,130. Total 5 integers. Don't forget about 1 and 130, they are also positive integers.

2. D

The product of all odd numbers will be odd number, so we exclude 0 even integers.

2 even and 2 odd integers and in the group. The product of odd integers will be odd number, but the product of even and odd numbers is an even integer and the product of even numbers is an even integer. Sum of 2 odd numbers is an even number, even + even=even.

4 even integers in the group. The product and the sum of all even numbers is even.

3) E.

x/3 should be divisable by 4,6 and 8.

[karmaholic]

Q1: Factorize 130. Factors are 1,2,5,10,13,26,65,130..so we have 8 positive integer values for x.

 

Q2: D. II and III. i.e .. the four numbers can have 2 or 4 even numbers.

 

s={2,4,3,5}. Product = 120. Sum = 14

p={2,4,6,8}. Product = 384. Sum = 20

 

Q3: E 72. 72 is divisible by 3 and the length of 3 equal parts = 24.

 

This 24 is the LCM of 4,6 and 8. (since 4,6 and 8 yeild integer length).

[Ankost]

Q1. Agree with A.

Stupid mistake.

[CTG1983]

Q.1 First find how many factors does it have. prime factorisation of 130 is 2*5*13. So no of factors is 8. Ans. A

[bragss2]

nice approach ctg

[Bright]

OA is a, d, e respectively................

[2005buster]

Q1: If N = (p^a) + (q^b) + (r^c). [p,q,r = prime factors, a,b,c = index powers]

 

I thought no of factors of N = (a+1) + (b+1) + (c+1).

130 = 2^1 + 5^1 + 13^1. Hence no of factors of 130 = 6 (although its 8)

I seem to be missing something somewhere. Pls assist.

[karmaholic]

I seem to be missing something somewhere. Pls assist.

 

You are missing out on "1" and "130" which divide 130 completely.

[chix475ntu]

Q1: If N = (p^a) + (q^b) + (r^c). [p,q,r = prime factors, a,b,c = index powers]

 

I thought no of factors of N = (a+1) + (b+1) + (c+1).

130 = 2^1 + 5^1 + 13^1. Hence no of factors of 130 = 6 (although its 8)

I seem to be missing something somewhere. Pls assist.

Should be

N = (a+1) * (b+1) * (c+1).

[2005buster]

Yup! *, not +.

Thx Chix.