Question

find the smallest whole no. whose first digit is 4 and value of no. obtained by moving 4 to last place is 1/4 of its original value

Answer 1

Answer:

We know that b has a 1 in its last place so a must have a 7 in its last place because 7×3=21 is the only product of three which has a 1 in its last place. So

a=7 and b=7×3=21

Now the last digit of both numbers is correct. However, if a ends with 7 then b needs to end with 71 instead of 21. So we lack another 50. How can we get another 50 into b? By adding 50 to a! Because 5×3=15 is the only product of three which has a 5 in its last place. So

a=57 and b=57×3=171

Now the last two digits are correct. However, as above, if a ends with 57 then b needs to end with 571 instead of 171 so we need another 400. How can we get another 400 into b? By adding 800 to a! Because 8×3=24 is the only product of three which has a 4 in its last place. So

a=857 and b=857×3=2571

Answer 2

Answer:

The correct answer is 410256

Step-by-step explanation:

Assuming the complete question to be

Find the smallest whole no. whose first digit is 4 and value of no. obtained by moving 4 to last place is 1/4 of its original value

Options: a)410256 b)4564 c)43336 d)43332

Solution:
We are supposed to move the first digit to the last and then divide the original number by 4 and check whether we get the same number we obtained after shifting the first digit.

a) 410256 gives 102564. 410256/4=102564

b) 4564  gives 5644.  4564/4=1411

c) 43336 gives 33364. 43336/4=10834

d) 43332 gives 33324. 43332/4=10833

Therefore, the correct answer is 410256.