The last two digits of 2151415?a)81b)61c)91d)51Correct answer is optio...
The last two digits of the number 2151415 can be determined by finding the remainder when the number is divided by 100. This is because the remainder when dividing by 100 gives us the last two digits of a number.
To find the remainder, we can use the concept of modular arithmetic. The modulus is the number we divide by, which in this case is 100. We can take the modulus of a number by taking the remainder when dividing the number by the modulus.
Let's break down the given number and find its remainder when divided by 100:
2151415 ÷ 100 = 21514 with a remainder of 15
Therefore, the remainder is 15. This means that the last two digits of the number 2151415 are 15.
However, the options given are in the form of two-digit numbers. So, we need to find the remainder when 15 is divided by 100 to get a two-digit number.
15 ÷ 100 = 0 with a remainder of 15
Therefore, the remainder is 15. This means that the last two digits of the number 2151415 are 15. However, 15 is a one-digit number, so we need to add a zero before it to make it a two-digit number.
Hence, the correct answer is option 'D', 51.
The last two digits of 2151415?a)81b)61c)91d)51Correct answer is optio...
Unit digit of this expression is always 1 as the base ends with 1.
For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.
i.e, tenth place digit in 2151 is 5 and
tenth place digit in power 415 is 1
And the units digit in the product of 5 x 1 = 5
Therefore, last two digits of 2151415 is 51.