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16th KVJ Solutions2

Here you found the solutions of  16th KVJ Solutions2 for part a in detail explanation.

2Q) a) Determine the smallest positive integer x, whose last digit is 6 and if we erase this 6 and put it in left most of the number so obtained, the number becomes 4x.

Solution: We have two methods of solving this .

First Method:

Single digit number cannot satisfy the requirements as 6 cannot be equal to 4*6=24.

Let us check for two digit number.

Two digit number x = 10a+6.

Erase 6 and put in left most then we get 10*6+a =60+a



6*10-24 =39a ......(1) , a is a digit and cannot be in fraction.

So we cannot get any 2 digit number which satisfy the requirements.

Let us check for 3 digit number

x=100a+10b+6, erase 6 and write in left most , then we get 100*6+10a+b,


6*10^2-24=390 a + 39 b

6*10^2 -24=39(10a+b)......(2)

By this we get pattern  for n-digit number

6*10^n -24 =39((n-1)digit number)

6*10^n - 24 is 39 multiple. Since n-1 digit number is an integer.

2*10^n -8 is multiple of 13 (divide by 3)

2(10^n -4) is multiple of 13

2 is not a multiple of 13 , so 10^n -4 is a multiple of 13.

10^n=4 (mod 13) , find n.

for n=5 10^5 -4 =13*7692,

n=5 . Then the number is

6*10^5 -24 =39(number without 6 )

15384 =number without 6 in ones place

Therefore the required number  is 153846.



6 erased and put in left most of the number, then we get

615384, This must be equal to 4x

4x=4(153846) =615384.

Hence smallest number which satisfies the given requirements is 153846.

16th KVJ Solutions2 , Second method.

Let x be the number whose ones digit is 6 and satisfies all requirements.

It is of the form 10k+6

Therefore 4x =4(10k+6)

                        = 40k+24


Hence second last digit is 4

x is of the form 100m+46  since 4 is second last digit and 6 is last digit of x.

4x =4(100m+46)



Therefore third last digit of x is 8 .

Now  x is of the form 1000n+846    since 8, 4, 6 are the third , second and last digits respectively .


     =4000n + 3384


Hence 4th last digit is 3 .

x is of the form 10000p+3846




hence 5th last digit of x is 5 and repeated 4p+1 form so we can stop here .

Therefore x is of the form = 10000q+5384

take q=1 smallest digit (leading place cannot be 0)


Hence the number required is 153846 .

Follow any one of the method given to you for 16th KVJ Solutions2.

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