You've gotta be kidding me

Note that the number is equal to $\dbinom{7}{0}1000^0+\dbinom{7}{1}1000^1+...+\dbinom{7}{7}1000^7=(1000+1)^7=7^7\times11^7\times13^7$ , which has $(7+1)(7+1)(7+1)=\boxed{512}$ factors.

1 7 21 35 35 21 7 1 is the 7th row of Pascal's triangle. Any number 10^n+1 raised to a power m, will result in the nth row of Pascal's triangle represented as a number, each number in the triangle getting m digits. (Of course, if the nth row has a number > m digits, this pattern messes up). This can be verified by using a calculator with numbers 11, 101, 1001 etc. Simple multiplication shows us intuitively why that happens. Expanding using the binomial theorem gives formal justification.

Since this is the 7th row of the triangle, represented with 3 digits per number, this is (10^3+1)^7 = 1001^7. 1001 = 11 * 91 = 11 * 13 * 7

This number has exactly 3 prime factors, 7,11 and 13. To calculate the total number of positive divisors, we note that 7^7 11^7 13^7 can result in 8 8 8 = 512 different factors (including 1 and the number itself)

I used PAscal,but this is also quick solution:

z=1007021035035021007001=35x1001x10^9+21x(10^9+1)x10^6+7x(10^15+1)x10^3+10^21+1

Notist that 10^(3n)+1=(1000)^n+1=(-1)^n+1 (mod 1001)

so for odde n number 10^(3n)+1 is divided by 1001

Notis now that 999x1001x1000+1001=999999000+1001=10^9+1

so (10^9+1)/1001=999x1000+1=999001

10^15+1=(10^9+1)x10^6-10^6+1=1001x(999001x10^6-999)=999000999001

also 10^21+1=(10^15+1)x10^6-10^6+1=1001x(999000999001x10^6-999)=999000999000999001

so a=z/1001=1006015020015006001 = 15x(10^6+1)x10^6+6x(10^12+1)x1000+10^18+1+20x(10^9+1)-20

10^(6n)+1=(1000^2)^n+1=(-1x-1)^n+1=2 (mod 1001)

so a=15x2x1+6x2x(-1)+2-20=0 (mod 1001)

so b=a/1001=1005010010005001=5x(10^9+1)x1000+10^15+1+ 1001x10^7 is divided bu 1001 by previous

so c=b/1001=1004006004001=10^12+1+4x(10^6+1)x1000 is divided by 1001 by previous (2+2x1000 mod (1001))

so d=c/1001=1003003001=3x1001x1000+10^9+1 i sdivided by 1001 by previous

so e=d/1001=1002001=1001x1000+1001=1001x1001

now we see that z=1001^7=(7x11x13)^7= 7^7x11^7x13^7

so number of divisors is (7+1)^3=512

This was actually a problem from the HMMT guts round :)

Try test of divisibility try for 7, 11, 13 combined test make groups of 3 digits each from the end

1 007 021 035 035 021 007 001

add groups odd places 001 + 021 + 035 + 007 = 64

add groups even places 007 + 035 + 021 + 1 = 64

find difference if difference is zero then the given number is divisible by 7, 11 and 13

or multiple of either 7, 11, 13 the number is divisible by either of 7, 11, or 13

here difference is zero so it is divisible by 7, 11, and 13 i.e. by 1001

find factors by dividing by 1001 you will get it is 1001^7

so factors are 7^7 × 11^7 × 13^7

so number of divisors = (7 + 1)^7 - 512

Same method..:-)

1 7 21 35 35 21 7 1 is the 7th row of Pascal's triangle. Any number 10^n+1 raised to a power m, will result in the nth row of Pascal's triangle represented as a number, each number in the triangle getting m digits. (Of course, if the nth row has a number > m digits, this pattern messes up). This can be verified by using a calculator with numbers 11, 101, 1001 etc. Simple multiplication shows us intuitively why that happens. Expanding using the binomial theorem gives formal justification. Since this is the 7th row of the triangle, represented with 3 digits per number, this is (10^3+1)^7 = 1001^7. 1001 = 11 * 91 = 11 * 13 * 7 This number has exactly 3 prime factors, 7,11 and 13. To calculate the total number of positive divisors, we note that 7^711^713^7 can result in 888 = 512 different factors (including 1 and the number itself)