13. If the first 99 natural numbers are written side by side to form a new number 123456..........9899, then find the remainder when this number is divided by 11.
a. 5
b. 6
c. 7
d. 8
Answer: C
Explanation:
Any number that is divided by 11 leaves remainder which is equal to the difference of sum of digits in odd places and Sum of the digits in even places.
Sum of the digits at odd places: 12345678910111213141516171819.............96979899
= (1+3+5+7+9) + (0+1+2+3+4+.....9) × 9 = 430
(From 11 to 99, 1 to 9 occurs 9 times)
Sum of digits at even places:
123456789101112131415161718192021 ...........96979899
= (2 + 4 + 6 + 8 ) + 1 × 10 + 2 × 10 ..........9 × 10 = 20 + 450 = 470
Difference = 470 - 430 = 40
So remainder = 40/11 = 7
14.123456789101112.... . 434445 . Find the remainder when divided by 45?
a. 0
b. 10
c. 20
d. 30
Answer: A
Explanation:
Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.
Let the given number is N.
N has unit digit 5 so it is divisible by 5.
Now the divisibility for 9 is sum of the digits of N should be divisible by 9.
Digit sum of N = (1 + 2 + 3 + ... + 9) + 1 × 10 + (1 + 2 + 3 + ... + 9) + 2 × 10 + (1 + 2 + 3 + ... + 9) + 3 × 10 + (1 + 2 + 3 + ... + 9) + 4 × 6 + (1 + 2 + 3 + 4 + 5)
= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 24 + 15 = 279
279 when divided by 9, remainder 0. So N can be written as 9m
So, N = 5k = 9m
So the given N is both multiple of 5 and 9. So it is exactly divisible by 45. So remainder = 0
15. Find the value of $\displaystyle\frac{{LCM{\rm{ (1, 2, 3, 4,}}.........{\rm{100)}}}}{{LCM{\rm{ (51, 52, 53, }}......{\rm{100)}}}}$
Answer: 1
Explanation:
LCM is defined as the product of all the prime numbers with maximum powers in the given numbers. There are 25 primes below 100 and we need to consider all the prime number where they take their maximum power.
LCM of (1, 2, 3, .........100) = ${2^6} \times {3^4} \times {5^2} \times {7^2} \times 11 \times 13.............97$
so LCM of (51, 52, 53, ...........100) = ${2^6} \times {3^4} \times {5^2} \times {7^2} \times 11 \times 13.............97$
For example 2 takes its maximum power 6 in 64. i.e., ${2^6}$. Similarly 3 takes its maximum power 4 in 81. i.e., ${3^4}$. so
Now to find the LCM of (51, 52, 53......100) we need to consider the prime numbers up to 100 as 100 is the maximum number. Again we can find the maximum power of 2 is 6 in 64. for 3 it is 4 in 81. for 5 it is 2 in 75 or 100...
$\displaystyle\frac{{LCM{\rm{ (1, 2, 3, 4,}}.........{\rm{100)}}}}{{LCM{\rm{ (51, 52, 53, }}......{\rm{100)}}}}$ = $\displaystyle\frac{{{2^6} \times {3^4} \times {5^2} \times {7^2} \times 11 \times 13.............97}}{{{2^6} \times {3^4} \times {5^2} \times {7^2} \times 11 \times 13.............97}}$= 1
16. Find the number of combinations of (a, b, c) if LCM (a, b) = 1000, LCM (b, c) = 2000, LCM (c, a) = 2000
a. 70
b. 80
c. 90
d. 100
Calculation for powers of 2:
$\text{LCM(c, a)}$$=\text{LCM}({2^r} \times {5^u},\, {2^p} \times {5^s})$ $={2^4} \times {5^3}$ . So one of $p$ or $r$ will take $4$. If $p$ is $4$ then, power of $2$ in $\text{LCM(a, b)}$ will be $4$. But power of $2$ is only given $3$. So $r= 4$. Either $p$ or $q$ will take 3 and other will take 0, 1, 2, 3.
$r = 4;\; p = 3;\; q = 0, 1, 2, 3$ if $p$ takes 3
$r = 4;\; p = 0, 1, 2;\; q = 3$ if $q$ takes 3
Total combinations are $(4, 3, 0), (4, 3, 1), (4, 3, 2), (4, 3, 3)$ and $(4, 0, 3), (4, 1, 3), (4, 2, 3)$.
Total options for powers of 2 = 7
Calculation for powers of 5:
Maximum power of 5 is 3. So any 2 of $s, t, u$ have maximum power 3. and other will take 0, 1, 2, 3
$s = 3;\; t = 3;\; u = 0, 1, 2, 3$ if $(s, t)$ are $3$
$s = 0, 1, 2;\;t = 3;\; u = 3$ if $(t, u)$ are $3$
$s = 3;\; t = 0, 1, 2;\; u = 3$ if $(s, u)$ are $3$
Total options are $7 \times 10 = 70$
17. The letters A, B, C, D, E, F and G represent distinct digits chosen from (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) such that A*B*C = B*G*E = D*E*F, where ‘*’ means multiplication. What does the letter G represent?
a. 4
b. 3
c. 2
d. 5
Answer: A
Explanation:
Assume $A \times B \times C = B \times G \times E = D \times E \times F = X$ then A, B, C, D, E, F, G all divides $X$ exactly. That means, $X$ is the LCM of all these digits. $\text{LCM}$ of $(1, 2, 3, 4, 6, 8, 9) = 72$
As $A \times B \times C = B \times G \times E = D \times E \times F $ - - - - (1), so each expression must posses those numbers whose integral multiple or sub-multiples is/are possessed by the other expressions. Thus 0, 5 & 7 are ruled out.
But $72 = 1 \times 8 \times 9 = 2 \times 4 \times 9 = 6 \times 4 \times 3$
Here 4, 9 appeared two times so these values take B, E in some order so $G = 2$
We can't determine the remaining values uniquely.
18. Find the remainder when 39! is divided by 41.
a. 1
b. 11
c. 39
d. 40
Answer: A
Explanation:
Substituting p = 41 in the wilson's theorem, we get
\(\dfrac{{40! + 1}}{{41}} = 0\)
\(\dfrac{{40 \times 39! + 1}}{{41}} = 0\)
\(\dfrac{{ - 1 \times 39!}}{{41}} = - 1\)
Cancelling -1 on both sides,
\(\dfrac{{39!}}{{41}} = 1\)
Alternatively:
By using congruent method
$(41 - 1)! + 1 \equiv 0\left( \text{mod 41} \right)$
40! + 1 = 0 (mod 41)
40 × 39! = -1 (mod 41)
- 1 × 39! = -1 (mod 41)
Cancelling -1 on both sides,
39! = 1 (mod 41)
So the remainder when 39! is divided by 41 is 1.
a. 4