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11. What is the sum of all two digit numbers that gives a remainder of 3 when they are divided by 7?

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Solution:
The two digit number which gives a remainder of 3 when divided by 7 are:
10, 17, 24 ..... 94.
Now, these number are in AP series with
1st Term, a = 10;
Number of Terms, n = 13;
Last term, L = 94 and
Common Difference, d = 7.
Sum,
={n×a+L2}=13×52=676
12. The remainder , when (1523 + 2323) is divided by 19, is

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Solution:
NOTE: an + bn is always divisible by (a + b) when n is odd.

So, (1523 + 2323) always divisible by 38.
And 38 is a multiple of 19, So, the number which is divisible by 38, is divisible by 19 too.
(1523 + 2323) is divisible by 19.
So, 1523+232319   == Remainder ⇒ 15+2319  == Remainder ⇒ 3819 == Remainder ⇒ 0
13. A heap of pebbles when made up into group of 32, 40, 72, leaves the remainder 10, 18 and 50 respectively. Find least number of pebbles in the heaps.

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Solution:
In this type of problem we find the difference of divisors and their remainders.
Here difference,
32 - 10 = 22
40 - 18 = 22
72 - 50 = 22
Here, in each case difference is same i.e. 22
Then required number of pebbles is given by
[(LCM of 32, 40, 72) -22]

32 = 2 × 2 × 2 × 2 × 2
40 = 2 × 2 × 2 × 5
72 = 2 × 2 × 2 × 3 × 3
Hence,
LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440
Thus,
Required number of pebbles,
= 1440 - 22
= 1418
14. The sum of two numbers is 684 and their HCF is 57. Find all possible pairs of such numbers.

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Solution:
Since, HCF of two numbers = 57
Hence, numbers are multiples of 57
Let, the numbers be 57x and 57y, where x and y are prime to each other.
According to the sum,
57x + 57y = 684
Or, x + y = 12
Hence, required possible pair of values of x and y which are prime to each other are (1, 11) and (5, 7).
Thus, required pairs of numbers are,
{57 × 1 = 57 and 57 × 11 = 627}

{57 × 5 = 285 and 57 × 7 = 399}
15. Two numbers are in the ratio 3 : 4. If their LCM is 240, the smaller of two number is

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Solution:
Let, these two numbers be 3x and 4x then their LCM = 12x
Now, according to question,
12x = 240
Or, x = 20
Thus, the numbers are (3x = 3 × 20) = 60 and (4x = 4 × 20) = 80
Then smaller in this two is 60
16. The HCF of two numbers, each having three digits, is 17 and their LCM is 714. The sum of the numbers will be:

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Solution:
Let the numbers be 17x and 17y where x and y are co-prime.
LCM = 17xy
Now, 17xy = 714
Or, xy = 42 = 6 × 7
→x = 6 and y = 7
Or, x = 7 and y = 6
1st number = 17 × 6 = 102
2nd number = 17 × 7 = 119
Sum = 102 + 119 = 221
17. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32, 35 is.

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Solution:
LCM of 20, 28, 32, 35 will be the greatest number which is divisible by these numbers.
Firstly, we find LCM of 20, 28, 32, 35
20 = 2 × 2 × 5
28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
35 = 5 × 7
LCM = 2 × 2 × 2 × 2 × 2 × 5 × 7 = 1120
Required greatest number which subtract from 5834 are divide by 20, 28, 32 and 35
= 5834 - 1120
= 4714
18. Which among 212, 313, 414, 616 and 12112 is largest?

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Solution:
LCM in power 2, 3, 4, 6, 12 is 12
We multiplied the LCM to the power of the numbers.
21×122,31×123,41×124,61×126and121×1212
We get,
= 26, 34, 43, 62, 12
= 64, 84, 64, 36, 12
Hence, greatest number would be 313
19. LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is

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Solution:
Let two numbers be N1 and N2.
Now, we have
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
936 × 4 = 72 × N1
Or, N1 = 936×472
Or, N1 = 52
20. The rightmost non-zero digit of the number 302720 is

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Solution:
(30)2720, we can write it as[(30)4]680
Or,[(10 × 3)4]680
The right most non-zero digit depends on the unit digit of [(3)4]680
Unit digit of [(3)4]680,
Or, (81)680
The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1
Thus, required unit digit is 1

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