# congruences

how to do this q using congruences

find the remainder wen 32n+2-8n-9 is divided by 64?

i know the procedure by binomial

3
Swami Dayal ·

can we do this with congruences???if so how?

2
Ersin ·

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• Ersin This is just an attempt tp put something on
2
Ersin ·

Any natural number n is one of the forms 8k , kâˆˆN, 8k+1, 8k+2, 8k+3, 8k+4, 8k+5, 8k+6, 8k+7, kâˆˆ N âˆª{0}.

Let S(n)=9n+1 -8n-9. Let' s prove it 64|S(n), where n is any of the mentioned forms, which means that 64|S(n) for any natural number n.

Let' s first notice that it is

91 â‰¡ 9 (mod 64); 92=81 , 81 â‰¡ 17 (mod 64), 92 â‰¡ 17 (mod 64),

93â‰¡ 9Ã—17 (mod 64); 9Ã—17 =153 , 153 â‰¡ 25 (mod 64), 93 â‰¡ 25 (mod 64),

94 â‰¡ 9Ã—25 (mod 64); 9Ã—25 =225 , 225 â‰¡33 (mod 64), 94 â‰¡ 33 (mod 64),

95 â‰¡ 9Ã—33 (mod 64); 9Ã—33 =297 , 297 â‰¡41 (mod 64), 95 â‰¡ 41 (mod 64),

96 â‰¡ 9Ã—41 (mod 64); 9Ã—41 =369 , 369 â‰¡49 (mod 64), 96 â‰¡ 49 (mod 64),

97 â‰¡ 9Ã—49 (mod 64); 9Ã—49 =441 , 441 â‰¡57 (mod 64), 57â‰¡ -7 (mod 64), 97 â‰¡ -7 (mod 64),

98 â‰¡ -7Ã—9 (mod 64); -7Ã—9 =-63 , -63 â‰¡1 (mod 64), 98 â‰¡ 1 (mod 64).

We have

1Â° n=8k, kâˆˆNâˆª{0}. Then S(8k)=9Ã—98k-64k-9,

S(8k)â‰¡9Ã—98k-9 (mod 64), 9Ã—98k-9â‰¡9Ã—1-9 (mod 64),

S(8k)â‰¡0 (mod 64) (from 98â‰¡1 (mod 64) follows 98kâ‰¡ 1 (mod 64)),

2Â° n=8k+1, kâˆˆNâˆª{0}. S(8k+1)=9Ã—98k+1-64k-8-9,

S(8k+1)=92Ã—98k-64k-17, S(8k+1)â‰¡92Ã—98k-17 (mod 64),

92Ã—98k-17â‰¡17Ã—1-17 (mod 64), S(8k+1)â‰¡0 (mod 64),

3Â° n=8k+2, kâˆˆNâˆª{0}. S(8k+2)=9Ã—98k+2-64k-16-9,

S(8k+2)=93Ã—98k-64k-25, S(8k+2)â‰¡93Ã—98k-25 (mod 64),

93Ã—98k-25â‰¡25Ã—1-25 (mod 64), S(8k+2)â‰¡0 (mod 64),

4Â° n=8k+3, kâˆˆNâˆª{0}. S(8k+3)=9Ã—98k+3-64k-24-9,

S(8k+3)=94Ã—98k-64k-33, S(8k+3)â‰¡94Ã—98k-33 (mod 64),

94Ã—98k-33â‰¡33Ã—1-33 (mod 64), S(8k+3)â‰¡0 (mod 64),

5Â° n=8k+4, kâˆˆNâˆª{0}. S(8k+4)=9Ã—98k+4-64k-32-9,

S(8k+4)=95Ã—98k-64k-41, S(8k+4)â‰¡95Ã—98k-41 (mod 64),

95Ã—98k-41â‰¡41Ã—1-41 (mod 64), S(8k+4)â‰¡0 (mod 64),

6Â° n=8k+5, kâˆˆNâˆª{0}. S(8k+5)=9Ã—98k+5-64k-40-9,

S(8k+5)=96Ã—98k-64k-49, S(8k+5)â‰¡96Ã—98k-49 (mod 64),

96Ã—98k-49â‰¡49Ã—1-49 (mod 64), S(8k+5)â‰¡0 (mod 64),

7Â° n=8k+6, kâˆˆNâˆª{0}. S(8k+6)=9Ã—98k+6-64k-48-9,

S(8k+6)=97Ã—98k-64k-57, S(8k+6)â‰¡97Ã—98k-57(mod 64),

97Ã—98k-57â‰¡-7Ã—1-57 (mod 64), S(8k+6)â‰¡0 (mod 64),

8Â° n=8k+7, kâˆˆNâˆª{0}. S(8k+7)=9Ã—98k+7-64k-56-9,

S(8k+7)=98Ã—98k-64k-65, S(8k+7)â‰¡98Ã—98k-65(mod 64),

98Ã—98k-65â‰¡1Ã—1-65 (mod 64), S(8k+7)â‰¡0 (mod 64).

Therfor S(8k+j)â‰¡0 (mod 64), jâˆˆ{0,1,2,3,4,5,6,7} whence it follows S(n)â‰¡0 (mod 64), for any natural number n.

The task was done by Dr. DÅ¾enis PuÄiÄ‡ at State University of Novi Pazar, Serbia.