Question

when 534527 is divided by 1000 the remaimder is equal to​

Answer 1

Answer:

Without being clever, let's brute force our way through.

We want to understand 22014−1(mod1000)22014−1(mod1000). Notice that gcd(3,1000)=1gcd(3,1000)=1, so 33 has a modular inverse (a quick check shows that it's 667667).

Now, how do we find 22014−1(mod1000)22014−1(mod1000)? As gcd(2,1000)=2gcd(2,1000)=2, we should worry about the 22 factor. So we think of 10001000 as 23⋅5323⋅53.

Clearly 22014≡0(mod23)22014≡0(mod23). And as φ(53)=100φ(53)=100, we know that 22014≡214≡9(mod125)22014≡214≡9(mod125). Putting these together, perhaps with the Chinese Remainder Theorem or by quick brute force (there are only 8 things to check, afterall), we see that 22014≡384(mod1000)22014≡384(mod1000).

Thus 22014−1≡383(mod1000)22014−1≡383(mod1000). And thus 22014−13≡667⋅(22014−1)≡667⋅383≡461(mod1000)22014−13≡667⋅(22014−1)≡667⋅383≡461(mod1000).

So the remainder is 461461.

Answer 2

Answer:

The remainder is $527$ when the number $534527$ is divided by $1000$.

Step-by-step explanation:

Recall the divisional algorithm theorem,

$a=bq+r$, where $a=$ dividend, $b=$ divisor, $q=$ quotient and $r=$ remainder such that either $r=0$ or $r < b$.

Consider the dividend as follows:

$534527$

To find: The remainder when $534527$ is divided by $1000$.

Using the divisional algorithm, we have

$534527=1000\times 5+34527$

Here notice that the remainder is $34527 > 1000$.

Again, applying the divisional algorithm to $34527$ and $1000$, we get

$34527=1000\times 3+4527$

Here, the remainder is $4527 > 1000$.

Applying the divisional algorithm one more time to $4527$ and $1000$, we get

$4527=1000\times 4+527$

Here, the remainder is $527 < 1000$.

Therefore, the remainder is $527$ when the number $534527$ is divided by $1000$.

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