Question

The 6 digit number a9471b is divisible by 63.what is the value of a+b?

Answer 1

Answer:

100044 is the 6 digit number which is divisible by 63.

Answer 2

Answer:

a+b=6, number 294714

Step-by-step explanation:

• Given:

Number a9471b is divisible by 63

• Find: a+b

Number is divisible by 63=7*9 if it is divisible by both 7 and 9

Divisibility by 9 rule:

• Sum of numbers must be divisible by 9 => a+9+4+7+1+b= a+b+21 must be divisible by 9. It must be equal to 27 or 36 (because 0<a+b<18)

a+b+21=27 => a+b=6

a+b+21=36 => a+b=15

Divisibility by 7 rule:

• Take the digits of the number in reverse order, that is, from right to left, multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary. Then add the products. If the resulting sum is divisible by 7, then the original number is divisible by 7.

b1749a =>b+3+14+24+36+5a=5a+b+77

Since 77 is divisible by 7, 5a+b must be divisible by 7. If we put it as 5a+b=7k,

We have combination of conditions:

a+b=6 or a+b=15

and

5a+b=7k

Let's investigate these cases:

1) a+b=6 and 5a+b=7k

Here we need to consider: 1≤a<6 and 0<b≤6  (as a+b=6)

The difference of two equations gives us:

4a=7k-6 => 4a+6=7k => 2a+3=7m

Possible cases:

• a=1 => 5≠7m
• a=2 => 7=7m, then b=4
• a=3 => 9≠7m
• a=4 => 11≠7m
• a=5 => 13≠7m
• a=6 => 15≠7m
• One solution in this scenario

2) a+b=15 and 5a+b=7k

Here we consider: 6≤a,b≤9  (as a+b=15)

The difference of two equations gives us:

4a=7k-15 => 4a+15=7k

Possible cases:

• a=6 =>39≠7k
• a=7 => 43≠7k
• a=8 => 47≠7k
• a=9 => 51≠7k
• No solutions in this scenario.

So the only solution is a+b=6, a=2, b=4 and the number is  294714