Math, asked by Lavi7063, 4 months ago

Q3. Find the digits A and B in the number 235A11B so that it is divisible by 3 and when the digits A and B are interchanged, it becomes divisible by 8. *

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Answered by backfailure

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Answered by RvChaudharY50

Question :- Find the digits A and B in the number 235A11B so that it is divisible by 3 and when the digits A and B are interchanged, it becomes divisible by 8. ?

Solution :-

we know that,

if sum of all digits of a number is divisible by 3, then the number also divisible by 3.
if last three digits of a number are divisible by 8, then the number also divisible by 8 .

so,

→ (2 + 3 + 5 + A + 1 + 1 + B) ÷ 3 = Remainder 0

→ (12 + A + B) ÷ 3 = Remainder 0

→ (12/3) + (A + B)/3 = Remainder 0

→ (A + B)/3 = Remainder 0

Possible values of A and B are :-

A = 0, B = 3
A = 0, B = 6
A = 0, B = 9
A = 1, B = 2
A = 1, B = 5
A = 1 , B = 8
A = 2, B = 1
A = 2, B = 4
A = 2, B = 7
A = 3, B = 0
A = 3, B = 3
A = 3 , B = 6
A = 4, B = 2
A = 4, B = 5
A = 5, B = 1
A = 5, B = 4
A = 6, B = 0
A = 6 , B = 3
A = 7, B = 2
A = 8, B = 1
A = 9 , B = 0

now, given that, when the digits A and B are interchanged, it becomes divisible by 8.

so,

→ 235B11A ÷ 8 = Remainder 0

or,

→ 11A ÷ 8 = Remainder 0 .

putting values of A , we get,

if A = 0 , 110 ≠ Divisible by 8.
if A = 1 , 111 ≠ Divisible by 8.
if A = 2 , 112 = Divisible by 8.
if A = 3 , 113 ≠ Divisible by 8.
if A = 4 , 114 ≠ Divisible by 8.
if A = 5 , 115 = Divisible by 8.
if A = 6 , 116 ≠ Divisible by 8.
if A = 7 , 117 ≠ Divisible by 8.
if A = 8 , 118 ≠ Divisible by 8.
if A = 9 , 119 ≠ Divisible by 8.

therefore, we can conclude that,

A = 2.

hence,

if A = 2 , B can be = 1,4 or 7.