Problems on numbers :
In this section, we are going to see how problems on numbers can be solved.
Example 1 :
If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.
Let the number be ‘x’
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
Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.
Example 2 :
Find the number of prime factors of
610 × 717 × 5527
From 610 × 717 × 5527, we have to write each base in terms of multiplication of its prime factors.
The no. of prime factors = sum of the exponents
Hence, the number of prime factors is 91.
Example 3 :
Find the unit digit of
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
Hence the unit digit of the given number is 8.
Example 4 :
Find the square root of 123454321
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.
Hence the square root of 123454321 is 11111.
Example 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.
For each divisor and corresponding remainder, we have to find the difference.
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
Hence the required least number = 3465-10 = 3455
After having gone through the stuff given above, we hope that the students would have understood "Problems on numbers".
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