The 6 digit number a9471b is divisible by 63.what is the value of a+b?
Answers
Answer:
100044 is smallest 6 Digit number Exactly Divisible by 63
Step-by-step explanation:
Smallest 6 Digit number
= 100000
100000/63 = 1587 * 63 + 19
To be added 63 - 19 to make it divisible by 63
100000 + 63 - 19
= 100044
100044/63 = 1588
100044 is smallest 6 Digit number Exactly Divisible by 63
Given : The 6 digit number a9471b is divisible by 63
To find : Value of a + b
Solution:
6 digit number a9471b is divisible by 63.
Divisible by 63 means divisible by 7 & 9
Divisible by 9 if sum of digits of number is divisible by 9
=> a + 9 + 4 + 7 + 1 + b should be divisible by 9
=> a + b + 21 should be divisible by 9
=> a + b + 3 + 2(9)
a + b + 3 = 9 or 18
=> a + b =6 or a+b = 15
Divisible by 7 if
Divisibility rule for 7
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,
b(1) + 1(3) + 7(2) + 4(6) + 9(4) + a(5)
= b + 3 + 14 + 24 + 36 + 5a
= 5a + b + 77
=> 5a + b Should be divisible by 7
case 1 : a + b = 6
a , b = ( 1, 5) ,( 2, 4) , ( 3, 3) , ( 4 , 2) , ( 5 , 1) , ( 6 , 0)
only (2 , 4) Satisfied 5a + b divisible by 7
a = 2 & b = 4
case 2 : a+b = 15
(6 , 9) , ( 7 , 8) , (8 , 7) & (9 , 6)
None of the values satisfy 5a + b Should be divisible by 7
Hence value of a + b = 6
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