Math, asked by shauryasai371, 10 months ago

A number 40xyy129 is divisible by 33. Find the possible value x
y .also find the number​

Answers

Answered by amitnrw
0

x = 4  and y = 2 , 5 , 8  , Numbers  40422129  ,  40455129  & 40488129

Step-by-step explanation:

A number 40xyy129 is divisible by 33.

Divisible by 33

33 = 3 * 11

=> Divisible by 3 & 11

Divisible by 3 if sum of digit is divisible by 3

=> 4+ 0 + x + y + y + 1 + 2 + 9 is divisible by 3

=> x + 2y + 16  Divisible by 3

Divisible by 11

if subtract and then add the digits in an alternating pattern from left to right

and number got should be divisible by 11

4 - 0 + x - y + y - 1 + 2 - 9

= x - 4

=> x  = 4

x + 2y + 16

= 4 + 2y + 16

= 20 + 2y

y = 2  ,  5   ,  8  

x = 4

y = 2 , 5 , 8

Numbers

40422129

40455129

40488129

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