# Quantitative Reasoning – Practice – Number System: 6 quick questions! PSA_PRACTICE_QNR37. Solve the below questions:

1.The LCM of 84 and the number n is 588 and their GCF is 28.What is n?

2.What is the smallest integers greater than 2014 that is divisible by 101?

3.How many multiples of 16 are there between 100 and 1000?

4.If 7n is divided by 5,the remainder is 3.Find the remainder when 3n is divided by 5?

5.How many four digit numbers are multiples of 4 but not if 6?

6.What is the smallest positive integers which when multiplied to 2014 will make it a perfect square?

Ans:

1) For two natural numbers, their product = LCM * GCF.

Thus, n* 84 = 588 * 28
=> n =(588* 4)/12 = 196

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2) When 2014 is divided by 101, Quotient: 19 Remainder: 95

Thus, when (Divisor – Remainder) or (101-95) or 6 is added to 2014, the latter becomes divisible by 19.

Answer= 2014+6 = 2020.

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3)

16*7 = 112

16*62 = 992

Thus, there are 62-7+1= 56 multiples

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4)

When 7n is divided by 5,the remainder is 3; let us try finding a value of n that satisfies the condition.

We can observe that if n=4, the number = 28. Remainder is 3 when 28 is divided by 5.

Now, 3n= 3*4= 12. Remainder when 12 is divided by 5 = 2.

5)

Multiples of 4: 1000, 1004, 1008, 1012, 1016, 1020 …9996 ;

1000 = 4*250
9996= 4*2499

Total multiples= 2499-250+1= 2500-250 = 2250

The multiples we have to exclude are those which are not divisible by 6.

Among the multiples whichever are divisible by 3 would be divisible by 6. Looking at the first few numbers we must notice that every third number (1008, 1020,…) are divisible by 3.

Thus out of 2250 multilples, 1/3rd would be those which are divisible by 6.

Therefore, 2250 – (1/3)*2250= 2250 – 750 = 1500 are those which are multiples of 4 but not multiple of 6.

Ans ->1500

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6) Prime factorization of 2014 = 2* 1007 = 2* 19* 53

As all 2,9 and 53 are prime numbers, we would need another 2, 19 and 53 to be multiplied to make it a perfect square. Thus, smallest positive integer to be multiplied is 2014 itself!

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