Asked by dariu
I have some beads. When I divide all the beads into 4 equal groups, 1 bead is left. When I divide one such group into 4 equal groups again, the remainder will still be 1. Wen I split one such group into 4 equal groups again, the remainder is still 1. What is the minimum number of beads I have?
Answers
Answered by
mathhelper
4n+1
--> 4(4n+1) + 1 = 4n + 5
---- >4(4n+5) + 1 = 16n + 21
----> 4(16n + 21) + 1 = 64n + 84 + 1 = <b>64n + 85</b>
let n = 0
we get 85
testing:
85 ÷ 4 = 21 , rem 1
21 ÷ 4 = 5 , rem 1
5 ÷ 4 = 1 , rem 1
85 looks good
--> 4(4n+1) + 1 = 4n + 5
---- >4(4n+5) + 1 = 16n + 21
----> 4(16n + 21) + 1 = 64n + 84 + 1 = <b>64n + 85</b>
let n = 0
we get 85
testing:
85 ÷ 4 = 21 , rem 1
21 ÷ 4 = 5 , rem 1
5 ÷ 4 = 1 , rem 1
85 looks good
Answered by
Raghu Palisetty
--> 4(4n+1) + 1 = 4n + 5
could you please explain this step.
Regards
Raghu
could you please explain this step.
Regards
Raghu
Answered by
Raghu Palisetty
Your answer should substitute n = 1 and your equation was wrong in the first 2 statements ...
4a + 1 = Total beads
4b + 1 = a
4c + 1 = b
replace b with C in second equation 4(4c + 1 ) + 1 = a
16c + 5 = a
replace a with c in statement 1
4(16c + 5) + 1 = a.
64c + 21 = a.
Minmum number for c = 1.
then a = 85
4a + 1 = Total beads
4b + 1 = a
4c + 1 = b
replace b with C in second equation 4(4c + 1 ) + 1 = a
16c + 5 = a
replace a with c in statement 1
4(16c + 5) + 1 = a.
64c + 21 = a.
Minmum number for c = 1.
then a = 85
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