When a natural number ‘a’ is divided by another natural number ‘b’, if the quotient is ‘t’ and remainder is ‘r’ then ‘a’ can be uniquely expressed as a = tb + r where 0 LaTeX: \le ≤ r LaTeX: < < b , a > b. Let a = 666………6 consisting of 666 sixes. Then....
For b = 50, the value of r is_____
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Given : a = tb + r a = 666………6 consisting of 666 sixes. b =50
To find : Value of r
Solution:
a = tb + r
a = 666,..............6 Consisting of 666 times 6
Last two Digit end with 66
66 = 50 * 1 + 16
Hence r = 16
666.........6 666 times
= 666.....600 + 66 ( 6 664 times)
666.....600 = 666......6 * 100 ( 6 664 times)
is completely Divisibly by 50 ( as 100 is Divisble by 50)
66 = 50 * 1 + 16
Hence r = 16
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If a = bq +r the least value of r is - - Brainly.in
https://brainly.in/question/11962008
1. भाग लेमा a= bq+r में q= 0 तब होगा जब-(A)
https://brainly.in/question/13447180
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