Consider all positive integers between 3000 and 4000 which, when divided by any of the first three primes, leave a remainder of 1. Find the sum of all these integers.

Guest Dec 18, 2019

#1**+1 **

Haven't solved it yet...

though I have some ideas...

the first 3 prime numbers are 2,3,5. 2 would make the number between 3000 and 4000 odd. 5 has two options, 1 at the end of the number, or 6 at the end. 6 at the end is elimanated because of our first restriction. 3 would make the rest of the 3 digits sum up to 2, 5,8,11,14,17,20, 23, and 26. So the numbers that work have to be in the form abc1 where a+b+c=2, 5,8,11,14,17,20, 23, or 26. Thats my idea and I hope this helps.

Guest Dec 18, 2019

#2**+1 **

a=3;b=0;c=0;d=0;p=0; cycle:n=a*1000+b*100+c*10+d;if(n%2==1 or n%3==1 or d%5==1, goto loop, goto next); loop:printn," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<4, goto cycle,0);print"Total = ",p

OUTPUT:

There are 733 such numbers

Their total sum =2,565,566

P.S. The above code tests each number such as 3001 mod 2, or 3001 mod 3 or 3001 mod 5. If it has a remainder of 1, then it counts it and sums it up.

Guest Dec 18, 2019

edited by
Guest
Dec 18, 2019

#3**+1 **

Consider all positive integers between 3000 and 4000 which, when divided by any of the first three primes, leave a remainder of 1.

Find the sum of all these integers.

(3000+x) mod 2 = 1 therefore x is odd

(3000+x)mod 3=1

so

(3000+x) mod6 =1

(3000+x)mod5 =1

so

(3000+x)mod30=1

The smallest number that meets this requirement is x=1

the next is x=31,

then it goes up by 30.

The numbers will be

3000(1+31+61+ ........ 991)

Sum=3000*34+(1+(1+30)+(1+2*30)+ ........ (1+33*30))

Consider the AP a=1, d=30, n=34,

\(\qquad S_{n}=\frac{n}{2}(a+L)\\ \qquad S_{34}=\frac{34}{2}(1+991)\\ \qquad S_{34}=17*(992)\\ \qquad S_{34}=16864\\\)

**Sum= 3000*34+16864 = 118864 **

I have just reapaired my formulas in red. Now the answers to this interpretation of the question are the same.

Melody Dec 18, 2019

#6**+1 **

Hi Melody: Guest #4 comment is not me.

There are 1,000 numbers between 3000 and 4000. Their average value is 3,500 per number. So, their maximum sum should not exceed:3,500 x 1000 =3,500,000. Your total sum of 50 592 000 / 1 000 =50 592 per number. Hence the comment by Guest #4.

Take the first 20 numbers: 3001 3003 3004 3005 3006 3007 3009 3010 3011 3013 3015 3016 3017 3019 3021 3022 3023 3025 3026 3027. Each one of the above numbers when tested on mod 2, mod 3 and mod 5 leaves a remainder of 1, hence it must be counted and summed up. At least that is what I understand the question to be. Of course, I could be wrong too.

Guest Dec 18, 2019

#7**+1 **

I think that the question is badly worded.

The phrase " when divided by any of the first three primes ", is that to mean " if any one of them " or is it to mean " each one of them " ?

I'm with Melody's interpretation, that it means each one of them, (each one of them rather than any single one of them).

That gives rise to the sequence 3001, 3031, 3061, 3091, 3121, ... , 3991.

I don't agree with Melody's method of summation though.

(I get a sum of 118,864.)

Guest Dec 18, 2019

#8**+1 **

It is very likely that your interpretation and that of Melody are the accurate ones. Hence, there are only 34 such numbers as follows:

3001 3031 3061 3091 3121 3151 3181 3211 3241 3271 3301 3331 3361 3391 3421 3451 3481 3511 3541 3571 3601 3631 3661 3691 3721 3751 3781 3811 3841 3871 3901 3931 3961 3991 Total = 34

Which sum up to a total =118,864.

Guest Dec 18, 2019