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# Number System

1.

If n=1+x, where x is a product of four consecutive positive integers, then which of the following is true?

A. n is odd

B. n is prime

C. n is a perfect square

Since xx is the product of four consecutive integers, it is always divisible by 4, i,.e., it is always even. So, 1+x is always odd.

n=1+x

x=(y−1)(y)(y+1)(y+2)=y(y2−1)(y+2)=(y3−y)(y+2)=y4+2y3−y2−2y

⇒1+x=y4+2y3−y2−2y+1=y4+y2+1+2y3−2y2−2y=(y2+y−1)2

So, 1+x is a perfect square as we can see. Hence, option A is the correct choice.

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2.

How many three digit numbers can be formed using the digits 1,2,3,4,5,6,7 and 8 without repeating the digits and such that the tens digit is greater than the hundreds digit and less than the units digit?

Ten's digit = 7 ⇒ units digit =8 ⇒ Hundred's digit =1,2,3,4,5,6.
⇒ Number of ways =1×6

Ten's digit = 6 ⇒ units digit =7,8 ⇒ Hundred's digit =1,2,3,4,5.
⇒ Number of ways =2×5

Ten's digit = 5 ⇒ units digit =6,7,8 ⇒ Hundred's digit =1,2,3,4
⇒ Number of ways =3×4

Ten's digit = 4 ⇒ units digit =5,6,7,8 ⇒ Hundred's digit =1,2,3.
⇒ Number of ways =4×3

Ten's digit = 3 ⇒ units digit =4,5,6,78 ⇒ Hundred's digit =1,2
⇒ Number of ways =5×2

Ten's digit = 2 ⇒ units digit =3,4,5,6,7,8 ⇒ Hundred's digit =1
⇒ Number of ways =6×1

Total number of ways =6+10+12+12+10+6= 56.

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3.

How many ordered pairs of integer (x,y)(x,y) are there such that their product is a positive integer less than 100.

Given 0 x and y are either both positive or both negative.

Also given (x,y) is not equal to (y,x).
If x=1, y can take values from 1 to 99
⇒ we have 99×2=198 pairs but (1,1) is repeated
Thus can take 198−1=197 pairs.

If x=2 , yy can take values from 2 to 49
[(2,1) and (1,2) are also covered in 197 pairs above].
⇒ 48×2−1=95 pairs [(2,2) is repeated]

Similarly,
If x=3 or y=3 we have 61 pairs
If x=4 or y=4 we have 41 pairs
If x=5 or y=5 we have 29 pairs
If x=6 or y=6 we have 21 pairs
If x=7 or y=7 we have 15 pairs
If x=8 or y=8 we have 9 pairs
If x=9 or y=9 we have 5 pairs

We have total 473 pairs when x and y are positive.
We will have 473 pairs when a and b are negative.
⇒ We have a total of 946 ordered pairs.

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4.

A set of S consists of,

i). All odd numbers from 1 to 55.
ii). All even numbers from 56 to 150.

What is the index of the highest power of 3 in the product of all the elements of the set SS?

No answer description available for this question.

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5.

Three gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight. Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?

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6.

Find the remainder when PP is divided by 9?

We need to find sum of S of the digits of PP.
Till 999 each of the digits from 1 to 9 occurs equal number of times.

Total digit of all the numbers from 1 to 999 will be multiple of ∑9=45.

Hence,

S=45k+(1+0+0+0)
⇒ Required remainder = 1.

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7.

When 75% of a two-digit number is added to it, the digits of the number are reversed. Find the ratio of the unit's digit to the ten's digit in the original number.

No answer description available for this question.

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8.

The number 3 divides ‘a’ with a result of ‘b’ and a remainder of 2.

The number 3 divides ‘b’ with a result of 2 and a remainder of 1.

What is the value of a?

Since 3 divides b with a result of 2 and a remainder of 1, b=2×3+1=7.

Since number 3 divides a with a result of b (which we now know equals 7) and a remainder of 2, a=b×3+2=7×3+2= 23.

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9.

The highest power of 9 dividing 99! completely is

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10.

What is the remainder left after dividing 1!+2!+3!.....+100! by 7?

7!+8!+9!.......+100! is completely divisible by 7.

Now, 1!+2!+3!....+6!=873

When 873 is divided by 7 it leaves a remainder =5.