Mahindra Comviva Divisibility Problems with Solutions Quiz- 1

(1 – 3/7) * 3,80,205 = 2,17,260

When n is divided by 8, the remainder is 3 may be written as n = 8 k + 3 multiply all terms by 6 6 n = 6(8 k + 3) = 8(6k) + 18 Write 18 as 16 + 2 since 16 = 8 * 2. = 8(6k) + 16 + 2 Factor 8 out. = 8(6k + 2) + 2 The above indicates that if 6n is divided by 8, the remainder is 2. The answer is A

Three consecutive numbers can never be divisible by 3. The wording should have been “three consecutive multiples of 3”. Assuming that is what you meant the solution is given below.

Let the three numbers be (x-3), x and (x+3)

Their sum = 3x = 72 so x = 24.

So the three numbers are 21, 24 and 27.

The largest number is x+27.

The basic identity of a division is :

Dividend (number being divided) = Divisor x Quotient + Remainder

So, A = 7p + 3 and B = 7q + 5

Here p and q are the respective quotients (non-negative integer values).

Clearly, A - B = 7p + 3 - 7q - 5 = 7(p - q) - 2

Now on using the basic identity of a division, this equation implies that when (A - B) is divided by 7, the quotient is (p - q) and the remainder is -2.

However, the remainder (when dividing by 7) can only be integer values from 0 to 6. This means we need to make the remainder positive!

So we rewrite the above equation as: A - B = 7(p - q - 1 + 1) - 2

i.e. A - B = 7(p - q - 1) + 7 - 2

Hence, A - B = 7(p - q - 1) + 5

Again using the basic identity of a division, this means that when (A - B) is divided by 7, the quotient is (p - q - 1) and the remainder is 5.

The given number =7×86038 Sum of the odd places =8+0+8+7=23 Sum of the even places = 3+6+x (Sum of the odd places)- (Sum of even places) = Number (exactly divisible by 11) 23-(9+x) = divisible by 11 14 – x = divisible by 11. X must be 3, to make given number divisible by 11.

Explanation

Two-digit positive integers less than 100 are from 10 to 99. So, there are total 90 integers.

Suppose, U is the set of all integers from 10 to 99 and A, B, C are the sets of multiples of 2, 3, 5, respectively.

So, n(U)=90

n(A)= 90/2=  45

n(B)= 90/3= 30

n(C) = 90/5= 18

Number of integers which are divisible by both 2 and 3 : n(A ∩ B) = 90/6= 15

Number of integers which are divisible by both 3 and 5 : n(B ∩C) = 90/15 = 6

Number of integers which are divisible by both 2 and 5 : n(C∩ A) = 90/10= 9

and the number of integers which are divisible by all 2, 3 and 5 :

n(A∩ B∩ C) = 90/30= 3

n(A U B U C)

= n(A)+n(B)+n(C)-n(A∩ B)-n(B∩C)-n(C∩A) + n(A∩B∩C) =45+30+18-15-6-9+3 =66

So, the number of integers which are not divisible by 2, 3 and 5 will be:

n(U)- n( A U B U C)

= 90-66

=24

 

The simple answer of your question is 3 and 9.

Step-by-step explanation:

Hey Mate,

The simple answer of your question is 3 and 9.

For example,

Take any randomly 5 digit number , we take 12345.

so,

12345 - (1+2+3+4+5)

=> 12330

12330 is a multiple of 9 and 3 both.

Lets we take another example.

We take 11111

so,

11111 - (1+1+1+1+1)

=> 11106

11106 is a multiple of 9 and 3.

 

time taken= distance travelled / speed
time taken = 345/50
time taken=6.9hours

Here as per condition let us consider third number = x, then 2nd number = 2x
and 3rd number = 4x, Given that (4x+2x+x)/3 = 77, 7x = 3x77, x = 3 x 11 = 33,
first number = 4x = 4 x 33 = 132

So 9x/12x=9/12
time =distance /speed
4=360/9x
so x=10
12x=120
Time taken by car b =360/120=3 hours