Question

given a number x = 2^48-1 then between 5 and 10, x has


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Answer 1

Answer:

let N = 248−1 .

N is odd. So check all odd numbers between 5 and 10 i.e., 5,7,9.

Here a%b represents modulus operation i.e., value of remainder when a divided by b (a/b).

1) For 5 :

21=2,22=4,23=8,24=16,

25=32,26=64,27=128,28=256... and so on.

If power%4 = 0, unit digit will be 6.(eg. for 28 unit digit is 256)

So for 248 number unit digit = 6, and 248−1 number unit digit = 5.

So N is divisible by 5.

2) For 7 :

7 is factor, iff N%7=0 .

N=248−1≡0 (mod 7)

248≡1 (mod 7)

If 248%7=1 then 7 is a factor of N.

See fast exponentiation modulo algorithm for finding (ab)%m .

express 48 in binary = 110000

0 → 21%7=2

0 → 22%7=2∗2%7=4

0 → 24%7=4∗4%7=2

0 → 28%7=2∗2%7=4

1 → 216%7=4∗4%7=2

1 → 232%7=2∗2%7=4

Multiply all 1 terms (that are bold above)

248%7=2∗4%7=1

So 7 is a factor.

3) For 9 :

(explanation is same as above)

0 → 21%9=2

0 → 22%9=2∗2%9=4

0 → 24%9=4∗4%9=7

0 → 28%9=7∗7%9=4

1 → 216%9=4∗4%9=7

1 → 232%9=7∗7%9=4

Multiply all 1 terms (that are bold above)

248%9=7∗4%9=1

So 9 is a factor.

So 5,7,9 are factors of N.

Edit 1:

for checking 5 as factor, you can also use same method as shown above for 7.

If 248%5=1 then 5 is factor of N.

0 → 21%5=2

0 → 22%5=2∗2%5=4

0 → 24%5=4∗4%5=1

0 → 28%5=1∗1%5=1

1 → 216%5=1∗1%5=1

1 → 232%5=1∗1%5=1

Multiply all 1 terms (that are bold above)

248%5=1∗1%5=1

So 5 is a factor.

Answer 2

Step-by-step explanation:

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