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Test: Divisibility And Remainders- 3 - GRE MCQ


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10 Questions MCQ Test Quantitative Reasoning for GRE - Test: Divisibility And Remainders- 3

Test: Divisibility And Remainders- 3 for GRE 2024 is part of Quantitative Reasoning for GRE preparation. The Test: Divisibility And Remainders- 3 questions and answers have been prepared according to the GRE exam syllabus.The Test: Divisibility And Remainders- 3 MCQs are made for GRE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Divisibility And Remainders- 3 below.
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Test: Divisibility And Remainders- 3 - Question 1

What is the smallest integer that is multiple of 5, 7, and 20?

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 1

It is the LCM of 5, 7, and 20 which is 140. The answer is E.

Test: Divisibility And Remainders- 3 - Question 2

Which of these numbers is not divisible by 3? 

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 2

One may answer this question using a calculator and test for divisibility by 3. However we can also test for divisibilty by adding the digits and if the result is divisible by3 then the number is divisible by 3. 
3 + 3 + 9 = 15 , divisible by 3. 
3 + 4 + 2 = 9 , divisible by 3. 
5 + 5 + 2 = 12 , divisible by 3. 
1 + 1 + 1 + 1 = 4 , not divisible by 3. 
The number 1111 is not divisible by 3 the answer is D.

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Test: Divisibility And Remainders- 3 - Question 3

What is the smallest positive 2-digit whole number divisible by 3 and such that the sum of its digits is 9? 

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 3

Let xy be the whole number with x and y the two digits that make up the number. The number is divisible by 3 may be written as follows 
10 x + y = 3 k 
The sum of x and y is equal to 9. 
x + y = 9 
Solve the above equation for y 
y = 9 - x Substitute y = 9 - x in the equation 10 x + y = 3 k to obtain. 
10 x + 9 - x = 3 k 
Solve for x 
x = (k - 3) / 3 
x is a positive integer smaller than 10 
Let k = 1, 2, 3, ... and select the first value that gives x as an integer. k = 6 gives x = 1 
Find y using the equation y = 9 - x = 8 
The number we are looking for is 18 and the answer is D. It is divisible by 3 and the sum of its digits is equal to 9 and it is the smallest and positive whole number with such properties.

Test: Divisibility And Remainders- 3 - Question 4

 Let n! = 1 x 2 x 3 x……….x n for integer n> 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!),then p+2 when divided by 11! Leaves remainder of

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 4

If P = 1! = 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.

Test: Divisibility And Remainders- 3 - Question 5

The remainder, when (1523 + 2323) is divided by 19, is :

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 5

a+bn is always divisible by a + when n is odd.
Therefore 1523 + 2323 is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 1523 + 2323 is divisible by 19.
Therefore,the required remainder is 0.

Test: Divisibility And Remainders- 3 - Question 6

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 6

First of all, we have to identify such 2 digit numbers.
Obviously, they are 10, 17, 24, ….94
The required sum = 10 + 17 … 94.
Now this is an A.P. with a = 10, n = 13 and d = 7
Hence, the sum is

Test: Divisibility And Remainders- 3 - Question 7

What is the remainder when 1044 × 1047 × 1050 × 1053 is divided by 33?

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 7

You can solve this problem if you know this rule about remainders.
Let a number x divide the product of A and B.
The remainder will be the product of the remainders when x divides A and when x divides B.

Using this rule,
The remainder when 33 divides 1044 is 21.
The remainder when 33 divides 1047 is 24.
The remainder when 33 divides 1050 is 27.
The remainder when 33 divides 1053 is 30.

∴ the remainder when 33 divides 1044 × 1047 × 1050 × 1053 is 21 × 24 × 27 × 30.

Note:
The remainder when a number is divided by a divisor 'd' will take values from 0 to (d - 1). It will not be equal to or more than 'd'

The value of 21 × 24 × 27 × 30 is more than 33.
When the value of the remainder is more than the divisor, the final remainder will be the remainder of dividing the product by the divisor.
i.e., the final remainder is the remainder when 33 divides 21 × 24 × 27 × 30.
When 33 divides 21 × 24 × 27 × 30, the remainder is 30.

Test: Divisibility And Remainders- 3 - Question 8

If a positive integer n is divided by 5, the remainder is 3. Which of the numbers below yields a remainder of 0 when it is divided by 5?

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 8

n divided by 5 yields a remainder equal to 3 is written as follows
n = 5 k + 3 , where k is an integer.
add 2 to both sides of the above equation to obtain
n + 2 = 5 k + 5 = 5(k + 1)
The above suggests that n + 2 divided by 5 yields a remainder equal to zero. The answer is B.

Test: Divisibility And Remainders- 3 - Question 9

If an integer n is divisible by 3, 5, and 12, what is the next larger integer divisible by all these numbers?

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 9

If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and 12 which is 60.
n = 60 k
n + 60 is also divisible by 60 since
n + 60 = 60 k + 60 = 60(k + 1)
The answer is D

Test: Divisibility And Remainders- 3 - Question 10

If n is an integer, when (2n + 2)2 is divided by 4 the remainder is

Detailed Solution for Test: Divisibility And Remainders- 3 - Question 10

We first expand (2n + 2)2
(2n + 2)2 = 4n2 + 8n + 4
Factor 4 out.
= 4(n2 + 2n + 1)
(2n + 2)2 is divisible by 4 and the remainder is equal to 0. The Answer is A.

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