Question

answer it URGENT no. 8​

Answer 1

Explanation:

Let the missing numbers be a, b, c and d. So the equation is

$\mathsf{7a8b4\times7=5c1d58}$

To make things easier, we'll first write down the table for 7.

$7\times1=7, 7\times2=14,7\times3=21,7\times4=28, 7\times5=35, 7\times6=42,7\times7=49, 7\times8=56, 7\times9=63\ 7\times10=70$

First, $4\times7=28$. 8 is already in the first digit of the product and 2 is carried over to the next digit.

Next, we have $(b\times7)+2=n5$ (where n is the digit before 5).

This means $7b=n3$. The only number that satisfies the value of b is 9. (9 x 7 = 63).

∴ b = 9.

6 is carried over.

Next, we have $(8\times7)+6=nd$, meaning $62=nd$.

∴ d = 2.

6 is carried to the next digit again.

Then, we have $7a+6=n1$, meaning

$7a=(n-1)5$

The only value that satisfies a is 5, because then, (7 x 5) + 6 = 41.

∴ a = 5.

4 is carried over

Finally, we have $(7\times7)+4=nc$

$49+4=nc$

$nc = 53$

∴ c = 3, and 5 is given already.

Hence, the values are

a = 5, b = 5, c = 3, and d = 2, meaning the equation is

$75894\times7=531258$

Answer: a = 5, b = 5, c = 3, and d = 2

Hope this helps. :D