Lucky remainder

Since it has to leave a remainder of 7 thus is has to be divided by a number greater than 7..

We subtract 7 from the given number (2027-7=2020) and find all the factors of 2020 which are

1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020

But we take only those >7 i.e. 10, 20, 101, 202, 404, 505, 1010, 2020

Thus there are 8 numbers which leave remainder 7.

First, let us factorize $2027-7$ . This is $2^{2}*5*101$ . Now we look for numbers that can be obtained from these numbers that are larger than 7. They are 20, 101, 10, 505, 202, 404, 2020, and 1010, which is 8 numbers.

If 2027 divided by a, then leaves a remainder of 7... So, a must greater than 7 Then you can solve this with modulo, $2027 \equiv 7 \ mod a$ , hence $2020 \equiv 0 \ mod a$ from this, we know that a is divisor of 2020. there are 12 divisor of 2020. but 4 of them are less than 7. So, the possibilities of a is 8

If a number $K$ divided into $N$ leaves a remainder of $R$ , then $K$ has two important characteristics: $K | N-R$ and $K>R$ .

For this problem, $K=a$ , $N=2027$ , and $R=7$ . $a | 2027-7 \Rightarrow a | 2020$ and $a>7$ . The factors of $2020$ are $1$ , $2$ , $4$ , $5$ , $10$ , $20$ , $101$ , $202$ , $404$ , $505$ , $1010$ , and $2020$ . $8$ of these factors are greater than $7$ , so the answer is $\boxed{8}$ .

This problem translates to the modular equation $2027 \equiv 7 \pmod a \implies 2020 \equiv 0 \pmod a \implies a \mid 2020.$ So we want the number of positive integer divisors of $2020$ that are greater than $7$ .

We know that $2020 = 2^2 \times 5 \times 101$ , so it has $12$ positive integer divisors. But, of these, $1$ , $2$ , $4$ , and $5$ are bogus, as they are less than $7$ , so the answer is $12 - 4 = \boxed {8}.$

For $2027\%a = 7$ , $a$ has to be divisible by $2027-7 = 2020$ but not divisible by 2027

Divisors of 2020: 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020 Divisors of 2027 (prime no): 1, 2027

However, for $a$ to leave a remainder of $7$ , a>7 Therefore, the positive integer $a$ can be 10, 20, 101, 202, 404, 505, 1010, 2020 , which gives 8 numbers.

If $2027\equiv\ 7\mod{n}$ , then we immediately know that n must be a divisor of 2020. The prime factorization of 2020 is $2^2\times5\times101$ which makes the number of divisors equal to $3 \times 2 \times 2$ or 12 divisors. However, we are not done. If we divide 2027 by 1, the remainder is 0. So, we have to subtract any divisors of 2020 which are less than 7. This includes 1,2,4, and 5. So, from the total of 12 divisors we subtract 4 divisors to receive $\boxed{8}$ as the answer.

From division algorithm ,

Dividend = Divisor * Quotient + Remainder ( Divisor is strictly > Remainder and Remainder is strictly +ve)

Therefore ,

2027 = a * q + 7 ( where q = Quotient, some + ve integer )

2020 = a * q

Now ,

2020 = 1 * 2 * 2 * 5 * 101

and we also know that a>7.

Therefore , possilble values of a are

a = 10 , 20 , 101 , 202 , 404 , 505 , 1010 , 2020 .

So , a can have the above 8 + ve integral values.

We need $a \mid 2020 = 2^2 \cdot 5 \cdot 101$ , the number $2020$ has $3 \cdot 2 \cdot 2 = 12$ factors, but we need $a > 7$ , hence we ignore $4$ factors, namely $1,2,4,5$ , so there are $12 - 4 = \boxed{8}$ such values of $a$ .

2027-7=2020 2020=2^2x5^1x101^1 (2+1)(1+1)(1+1)=12 Factors 1, 2, 4 & 5 are under 7 so 12-4=8

All integers $n$ can be written in the form

$n=ar+b$

where $b$ is the remainder.

$2027=ar+7$

$2020=ar$

The prime factorization of $2020$ is $2^2\times5\times101$

So, the number of factors of $2020$ (which is the number of possible values of $a$ ) is $(2+1)\times(1+1)\times(1+1)=12$

However, in this problem $a>7$ , among the 12 possible values, $1, 2, 4, 5$ are not acceptable.

Thus, the answer is $12-4=8$

$a>7$ . there is a number p $(p>2027)$ such that $ap + 7 = 2027$ , $ap=2020$ . so a is a positif divisor of 2020 greater than 7, there are 8 numbers.

A simple C++ program can solve this

#include <iostream>
using namespace std;

int main() {
    int i,r,count=0;
    for(i=1;i<=2027;i++){
        r = 2027%i;
        if(r==7) count++;
    }
    cout<<count;
    return 0;
}

len([ i for i in range(1,2027) if 2027%i == 7])

First subtract 7 from 2027 which gives 2020. 2020= 2×2×5×101. Therefore total no of factors can be calculated using P&C =3×2×2=12, but 1,2,4&5 can't leave 7 as remainder so no. of factors of 2020 greater than 7 are 12-4=8

I wrote a program for it and got it write, took less than 2 minutes lol

an+7=2027, we need a>=8; there for, solve for n= 2020/a for:{1,2,3,4,5,6,7,8} as a values, we obtain n={ 2020, 1010,505,404,202,101,20,10}. (8 integers)

After reading the problem you should realize that; in order for a remainder of 7 to result the divisor MUST be greater than 7. So now we should doing the work.

Step 1: Rewrite the words into an equation we can solve. Let x be the other variable a is multiplied by. So the equation would be ax + 7 =2027

Step 2: Simplify the equation. ax = 2027-7 ax = 2020

Step 3: Find the factors of 2020. One way to ensure you found all of the factors is by starting off with prime factorization. The prime factorization of 2020 is 2020 = 2 \times 2 \times 5 \times 101 = 2^{2} \times 5 \times 101

Step 4 (3.5): Now multiply these to find all the factors of 2020. 2020 = 1 \times 2 \times 4 \times 5 \times 10 \times 20 \times 101 \times 202 \times 404 \times 505 \times 1010 \times 2020

Step 5: Get rid of the factors less than 7 (reasoning in first paragraph). So the factors you result with are 10, 20, 101, 202, 404, 505, 1010, 2020

Step 6: Conclusion - find final answer. For the answer just count up the number of numbers in the number set. Which is \boxed{8}

Given the number 2027 we can simplify this into which numbers are factors of 2020. Since 2020 leaves 2027 with a remainder of 7. Factoring out 2020 we get 2 \times 2 \times 5 \times 101, using the factor rule we add one to each of there powers resulting in 3 \times 2 \times 2=12 Hence there are 12 factors. We then find these 12 factors and they are 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, and 2020. Now 1,2,4,5 cannot possibly be factors since they are less than 7 therefore this leaves us with 8 factors (10, 20, 101, 202, 404, 505, 1010)

a must be divisible by $(2027-7)=2020$

$2020 = 2^{2} \times 5 \times 101$

Distinct prime factors are $2, 5, 101$

In a factor, count of number 2 can be 0, 1 or 2 - there are 3 ways.

Count of number 5 and 101 can be 0 or 1 - there are 2 ways for each.

So total number of factors are $2 \times 2 \times 3 = 12$

This includes 1 (when all factors are present 0 times)

Of them, $1, 2, 5, 2 \times 2 = 4$ these 4 factors which are less than 7.

Remaining 8 factors match our condition

Temos que: $D = d \cdot q + r$ , então:

$2027 = a \cdot q + 7 \\\\ a \cdot q = 2020 \\\\ a \cdot q = 2^2 \cdot 5 \cdot 101$

A quantidade de divisores é dada por:

$(2 + 1) \cdot (1 + 1) \cdot (1 + 1) = \\\\ 3 \cdot 2 \cdot 2 = 12$

Todavia, quatro deles são menores que 7, portanto,

$12 - 4 = \\\\ \boxed{8}$

Let quotient be x. Therefore 2027 is equal to [a.x] plus 7. ................as 7 is the remainder. So 2020 is equal to [a.x]. Therefore,factors of 2020 are : {1 . 2020}, {2 .1010} , {4 .505}, {5 .404}, {10 .202} and {20 .101} Therefore a can be 10,20,101,202,404,505,1010 and 2020 because dividing by 1, 2, 4 and 5 to 2027 won't bring remainder to 7!!!!

Therefore there are 8 values that satisfy the given condition.

Since 2020 is the largest possible value of a , all other values of a must give remainders of 0 when dividing 2020. Ergo, all values of a must be factors of 2020.

The factors of 2020 are 1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, and 2020. Note that only values larger than 7 can give 7 as the remainder, so the factors less than 7 are excluded.

This leaves 8 values: 10, 20, 101, 202, 404, 505, 1010, and 2020.

2027=ak+7 and a>7

ak=2020

ak=2*2*5*101

a can be

• 101, (ignoring 2,2,5 because each of them are <7)
• 2*5,2*101,5*101,
• 2*2*5,2*2*101,2*5*101,
• 2*2*5*101

so count =1+3+3+1=8

:)

You just have to see how many positive divisors of 2020 (which is 2027 - 7) greater than 7 there are. Since 2020 = 2^2 * 5 * 101, there are (2+1) (1+1) (1+1)=12 divisors of 2020. Since 1, 2, 4 and 5 are smaller than 7, there are 8 positive integrers a which leave a remainder of 7.