Hi Nancy Please see if the question is complete. I thing there is a typographical error , the length of RT ( 6cm ) is already mentioned in the ...

Hi Mayank ! We know, the remainder when f(x) = a + bc + cx^2 + dx^3 +...... is divided by ( x - a ) is f(a). ( Check NS - 4 Class Sheet ) Hence , S...

Hi Nilesh ! A = (26 - 1) = ((23)3 -1) = ( 23 - 1 ) ( 23 + 1 ) B = (29 - 1) = (23 -1) ( 23 + 1 + 26 ) HCF (A, B) = ( 23 - 1 ) We know, Produ...

Hi Nilesh .Let the numbers be 13a and 13 b. ( Where a and b are coprime ) LCM of 13a and 13b = 13ab 13ab = 78 ab = 6 a = 1 , b = 6 or a = 2 , b = 3 He...

Alternate Approach Last two digits of [1785^95 ] = Last two digits of 85^95 Last two digits of= 17^95 × 5^95 Last two digits of17 × [(17)²...

1st method Divide the number by 100 . Remainder will be last two digits of that number 1785^95 mod 100 Using Chinese Remainder Theorem 1785^95 ...

(a5 )^even always ends with 25 ( a5)^Odd => If a is odd then last two digits 75 and if a is even then , it ends with 25.

Hello Vidisha ! Take 2² common from each term , 2²( 1²+11²+111²+....1111....11²) mod 9 4(1+2²+3²+....+49²) mod 9 4 × 49×50×99/6 mod 9 = 6.

Hello Ritika ! We know any digit repeated 6 times is always divisible by 7. so 8888......written 48 and 99999.... written 48 times is always div...

Hello Yamin! Select all even numbers from 50 - 98 { 50 , 52 , 54 , ......., 98 } Hence , maximum value of n = 25

Hello Apoorva , Select all even numbers from 50 - 98 { 50 , 52 , 54 , ......., 98 } Hence , maximum value of n = 25

Hello Apoorva ! For a natural number N the number of factors of N below √N is equal to number of factors of N above √N . So if N has five facto...

Hello Apoorva ! Smallest number which has five distinct prime factors is 2 × 3 × 5 × 7 × 11 = 2310 . All three digit natural numbers are smaller...

Hello Ritika , We know , Product of an odd natural number and 5 always ends with the unit digit 5. So we can select 2^0 ( 1 way ) 3^0 , 3^1 , 3^...

Hello Apoorva! x⅓ + y⅓ = 4160⅓ a• (65)⅓ + b ( 65)⅓ = 4 • (65)⅓ a + b = 4 ( 0, 4) ( 4, 0) ( 1 , 3) ( 3 , 1) ( 2 , 2) 5 ordered pairs