Name: Prime Numbers From 1 to 100
Uploaded: 2019-07-19
Duration: 7 min 49 s
Description: The number 2 is the only even prime number. But, how do you find the prime numbers between 1 and 100? This video explains the easiest way to find prime numbers.

Question

Find the least number whose last digit is

7

and which becomes

5

times larger when this last digit is carried to the beginning of the number.

Accepted Answer

Let, the required number be $. . . g f e d c b a 7 .$

Since it is given that $5 (. . . g f e d c b a 7) = 7 . . . g f e d c b a . . . (i)$

Now, when we multiply $5$ by $7$ we get $5 \times 7 = 35$ and by comparing the one's digit of the numbers on both sides of the equation $(i)$ we get $a = 5 .$

Putting this value of $a$ back in the equation $(i)$ we have $5 (. . . g f e d c b 57) = 7 . . . g f e d c b 5 . . . (i i)$

Now, when we multiply $5$ and $57$ we get $5 \times 57 = 285$ and by comparing the ten's digit of the numbers on both sides of the equation $(i i)$ we get $b = 8 .$

On putting the value of $b$ in the equation $(i i)$ we get $5 (. . . g f e d c 857) = 7 . . . g f e d c 85 . . . (i i i)$

Now, when we multiply $5$ by $857$ we get $5 \times 857 = 4285$ and by comparing the hundred's digit of the numbers on both sides of the equation $(i i i)$ we get $c = 2 .$

On putting the value of $c$ in the equation $(i i i)$ we get $5 (. . . g f e d 2857) = 7 . . . g f e d 285 . . . (i v)$

Now, when we multiply $5$ and $2857$ we get $5 \times 2857 = 14285$ and by comparing the thousand's digit of the numbers on both sides of the equation $(i v)$ we get $d = 4 .$

On putting the value of $d$ in the equation $(i v)$ we get $5 (. . . g f e 42857) = 7 . . . g f e 4285 . . . (v)$

Now, when we multiply $5$ by $42857$ we get $5 \times 42857 = 214285$ and by comparing the ten thousand's digit of the numbers on both sides of the equation $(v)$ we get $e = 1 .$

On putting the value of $e$ in the equation $(v)$ we get $5 (. . . g f 142857) = 7 . . . g f 14285 . . . (v i)$

Now, when we multiply $5$ by $142857$ we get $5 \times 142857 = 714285$ and by comparing the lac's digit of the numbers on both sides of the equation $(v i)$ we get $f = 7 .$

Thus, we will get the same digits as we have obtained above.

Hence, the least required number is $142857 .$

Find the least number whose last digit is $7$ and which becomes $5$ times larger when this last digit is carried to the beginning of the number.

Important Questions on Number Theory

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