$a^n+b^n$ is always divisible by $a+b$ when n is odd.

$5^{23}+{23}^{23}$ is always divisible by $15+23=38.$ As 38 is a multiple of 1$9,{15}^{23}+{23}^{23}$ is divisible by $19.$

Hence we get a remainder of 0.

We know, 992250 = 2¹ x 3⁴ x 5³ x 7².
The value of a + b + c + d = 1 + 3 + 4 + 2 = 10
 

Using options, we find that sum of numerator and denominator or 97 : 84 is (97 + 84) = 181 which is a prime number. Hence, it is the appropriate answer.

$\left(a\right)10<n<1000$

Let n is two digit number

$n=10a+b\hat{a}{†}^{\prime }{P}_n=ab,S_n=a+b$

then, $ab+a+b=10a+b$

$\hat{a}{†}^{\prime }ab=9a\hat{a}{†}^{\prime }b=9$

There are 9 such numbers $19,29,33,....,99.$

Then let n is three digit number

$\hat{a}{†}^{\prime }n=100a+10b+c\hat{a}{†}^{\prime }{P}_n=abc,S_n=a+b+c$

Then $abc+a+b+c=100a+10b+c$

$\hat{a}{†}^{\prime }abc=99a+9b$

$\hat{a}{†}^{\prime }bc=99+9$

But the maximum value for $bc=81$

And RHS is more than $99.$ Hence, no such number is possible.

Sol:

e. 55

From the definition of "seed"?, it is clear that we have to count number of integers between 1 and 500, which are divisible by 9. The smallest is 9 and the largest is 495. $9\stackrel{\sim }{A}\_\_1=9$ and $9\stackrel{\sim }{A}\_\_55=495.$ Hence there are 55 such numbers.

LCM of (7, 12, 16) = 336. If we divide 1856 by 336 then remainder is 176. Since it is given that remainder in this condition is 4. Hence the least number to be subtracted = (176 - 4) = 172.

Alternate Approach:

In case you don't remember this method: go from answer options by subtracting the answer option from 1856 and divide by each of 7,12, and 16 to get a reminder of 4.

Ans:(a)

There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 numbers which are divisible by 3. Out of remaining there are 7 numbers which are divisible by 5. Hence numbers which are not divisible by 2, 3, 5 = (50- 17 - 7) = 26

Alternate Approach:

Taking the Euler's number approach:

If we take the prime factors of a number then we get all the numbers that are co-prime to that number, which means all the multiples of the prime factors are removed, using the same formula we can find out the numbers that are not divisible by a particular set of numbers.

100=29

As we don't want the number 2, 3, and 5 also to in the list remove these 3 numbers, hence option will be 26

Ans: (a)

Let one number be x, then second number will be $(x+4);$
Thus $:\frac{1}{x}+\left(\frac{1}{(x+4)}\right)=\frac{10}{21};$
$\frac{(x+x+4)}{\left(x\right(x+4\left)\right)}=\frac{10}{21};$
$\frac{(2x+4)}{x(x+4)}=\frac{10}{21};$
$x=3$

Elimination strategy:

By looking at the answer option as the sum of reciprocal of the numbers will have to give $\frac{10}{21}$The number has to be a factor or multiple of 21. So directly option (c) can be eliminated.

Now the numbers that we will have to check will be 3 & 7 , 1 & 5, 17 & 21, 21 &25

The only possibility from here is 3 & 7. Hence option (a).

The number 774958A96B is divisible by 8 if 96B is divisible by 8. And 96B is divisible by 8 if B is either 0 or 8. Now to make the same number divisible by 9 sum of all the digits should be divisible by 9. Hence (55 + A + B) is divisible by 9 if (A + B) is either 0 or 8 $\Rightarrow$ either A = 0 or B = 8, or A = 8 or B = 0. Since the number is divisible by both A and B. Hence A and B may take either values, i.e., 8 or 0.

This is a generalized question: the important word "ANY" can be used here. Let us solve the question for some prime numbers greater than 6 i.e., 7, 11, 13 and 17. If these numbers are divided by 6, the remainder is always either 1 or 5.