# PROBLEMS ON NUMBERS

Problem 1 :

If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

Solution :

Let x be the number.

Then x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’.

In the above sentence we have 296 is multiplied by the constant "k", 75 is added to that. In this form , we consider the number 75 as remainder when the number x is divided by 296.

We want to find the remainder when we divide the number "x" by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.

So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.

x = 37 ⋅ 8k + 37 ⋅ 2 + 1

x = 37 ⋅ (8k + 2) + 1

So, the remainder is ‘1’ when the number ‘x’ is divided by 37.

Problem 2 :

Find the number of prime factors of

610 ⋅ 717  5527

Solution :

From 610  717  5527, we have to write each base in terms of multiplication of its prime factors.

= (2 ⋅ 3)10  ⋅ (7)17 ⋅ (5 ⋅ 11)27

= 210 ⋅ 310 ⋅ 717 ⋅ 527 ⋅ 1127

The no. of prime factors  =  sum of the exponents

= 10 + 10 + 17 + 27 + 27

= 91

So, the number of prime factors is 91.

Problem 3 :

Find the unit digit of

24382643

Solution :

Step 1 :

Take the last two digits in the power and unit digit in the base. They are 43 and 2.

Step 2 :

Divide the last two digits of the power by 4 (in all the problems) and get the remainder. That is, when 43 is divided by4, the remainder is 3.

Step 3 :

Now this remainder 3 has to be taken as exponent of the unit digit 2. That is, 23 = 8.

So, the unit digit of the given number is 8.

Problem 4 :

Find the square root of 123454321.

Solution :

In the given number, we have the first five natural numbers in ascending order up to 5.

After 5, we have the first four natural numbers in descending order.

Whenever we have a number like this and we want to find square root, we have to replace each digit by 1, up to the digit where we have the first n natural natural numbers in ascending order.

So, in our number 123454321, we have to replace each digit by 1 up to 5. That is the square root of 123454321.

So, the square root of 123454321 is 11111.

Problem 5 :

Find the least number which when divided by 35, leaves a remainder 25, when divided by 45, leaves a remainder 35 and when divided by 55, leaves a remainder 45.

Solution :

For each divisor and corresponding remainder, we have to find the difference.

35 - 25 = 10

45 - 35 = 10

55 - 45 = 10

we get the difference 10 (for all divisors and corresponding remainders)

Now we have to find the L.C.M of (35,45,55) and subtract the difference from the L.C.M.

L.C.M of (35,45,55) = 3465

So, the required least number is

= 3465 - 10

= 3455

Problem 6 :

Two numbers add upto 18 and their product is 72. What are the two numbers?

Solution :

We ca assume x and y as two positive numbers.

Given : Two numbers add upto 18.

x + y = 18

y = 18 - x ----(1)

Given : The product of two numbers is 72.

xy = 72

Substitute y = 18 - x into the above equation.

x(18 - x) = 72

18x - x2 = 72

-x2 + 18x - 72 = 0

Multiply both sides by -1.

x2 - 18x + 72 = 0

Solve by factoring.

x2 - 6x - 12x + 72 = 0

x(x - 6) - 12(x - 6) = 0

(x - 6)(x - 12) = 0

x - 6 = 0  or  x - 12 = 0

x = 6  or  x = 12

 When x = 6,(1)----> y = 18 - 6y = 12 When x = 12,(1)----> y = 18 - 12y = 6

x = 6  and  y = 12

or

x = 12  and  y = 6

Therefore, the two numbers are 6 and 12.

Verification :

6 + 12 = 18

18 = 18

The product of the two numbers is 72.

12 = 72

72 = 72

Problem 7 :

The difference between two positive numbers is 3 and their product is 70. What are the two numbers?

Solution :

Let x and y be the two numbers.

Given : The difference between two positive numbers is 3

x - y = 3

x = y + 3 ----(1)

Given : The product of two numbers is 30.

xy = 70

Substitute x = y + 3 into the above equation.

(y + 3)y = 70

y2 + 3y = 70

y2 + 3y - 70 = 0

Solve by factoring.

y2 - 7y + 10y - 70 = 0

y(y - 7) + 10(y - 7) = 0

(y - 7)(y + 10) = 0

y - 7 = 0  or  y + 10 = 0

y = 7  or  y = -10

Since the numbers are positive, y can not be -10.

So, y = 7.

Substitute y = 7 into (1).

x = 7 + 3

x = 10

Therefore, the two numbers are 10 and 3.

Verification :

The difference between two positive numbers is 3.

10 - 7 = 3

3 = 3

The product of the two numbers is 70.

10  7 = 70

70 = 70

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