All Answers 358087 Answers

Converting to base 10, we get 1 + 4 + 23 + 50.  This is 79.

Converting back to base 3, 79 = 2221_3.

That looks right to me.  Try a different order, like 3p + p = 0 + 5p, or even 3p + 5p = p + 0.

Because when factor it, you get this:

1000! = 2^994 * 3^498 * 5^249 * 7^164 * 11^98 * 13^81 * 17^61 * 19^54 * 23^44 * 29^35 * 31^33 * 37^27 * 41^24 * 43^23 * 47^21 * 53^18 * 59^16 * 61^16 * 67^14 * 71^14 * 73^13 * 79^12 * 83^12 * 89^11 * 97^10 * 101^9 * 103^9 * 107^9 * 109^9 * 113^8 * 127^7 * 131^7 * 137^7 * 139^7 * 149^6 * 151^6 * 157^6 * 163^6 * 167^5 * 173^5 * 179^5 * 181^5 * 191^5 * 193^5 * 197^5 * 199^5 * 211^4 * 223^4 * 227^4 * 229^4 * 233^4 * 239^4 * 241^4 * 251^3 * 257^3 * 263^3 * 269^3 * 271^3 * 277^3 * 281^3 * 283^3 * 293^3 * 307^3 * 311^3 * 313^3 * 317^3 * 331^3 * 337^2 * 347^2 * 349^2 * 353^2 * 359^2 * 367^2 * 373^2 * 379^2 * 383^2 * 389^2 * 397^2 * 401^2 * 409^2 * 419^2 * 421^2 * 431^2 * 433^2 * 439^2 * 443^2 * 449^2 * 457^2 * 461^2 * 463^2 * 467^2 * 479^2 * 487^2 * 491^2 * 499^2 * 503 * 509 * 521 * 523 * 541 * 547 * 557 * 563 * 569 * 571 * 577 * 587 * 593 * 599 * 601 * 607 * 613 * 617 * 619 * 631 * 641 * 643 * 647 * 653 * 659 * 661 * 673 * 677 * 683 * 691 * 701 * 709 * 719 * 727 * 733 * 739 * 743 * 751 * 757 * 761 * 769 * 773 * 787 * 797 * 809 * 811 * 821 * 823 * 827 * 829 * 839 * 853 * 857 * 859 * 863 * 877 * 881 * 883 * 887 * 907 * 911 * 919 * 929 * 937 * 941 * 947 * 953 * 967 * 971 * 977 * 983 * 991 * 997

You see 5^249? That is 249 "5s" that are multiplied by 249 "2s". And: 5 x 2 =10 and you get 249 zeros at the end! And that is why.

First few digits
(10^15)!==117879641194089947860671229...

Number length
1.45657×10^16 decimal digits

Number of trailing zeroes
249999999999997

Last non-zero digits
...8442576510976

Could you please give us a link to the original question so that I can link the 2 together. Thanks ItsFree

Here is the pic.

Check it is right then try to answer from there.

Ask questions if you want.

You ask specific questions and learn from the answers.

To start with if you are adding you can think of  a number inside a square root as like a letter

For example.

Hopefull you know that    \(x+y \) can not be simplified

In the same way       \(\sqrt2+\sqrt3\)       cannot be simplified.

But hopfully you also know that      \(1x+5x+y=6x+y\)

Likewise          \(1\sqrt2+5\sqrt2+\sqrt3=6\sqrt2+\sqrt3\)

I guess that means you were only interested in the answer? 

THank you so much Melody. 

Thanks a lot. 

I have another question

q1. ∫(3x⁴-4)/ (2x³) dx

The equation 9^(x - 1)/3^(y + 2) = 81 can be rearranged to give 3^(y + 2) = 9^(x - 1). Taking the log of both sides, we have y + 2 = (x - 1)log(9). Therefore,

y = (x - 1)log(9) - 2

= 2(x - 1) - 2

= 2x - 2 - 2

= 2x - 4.

stop posting obvious homework questions and making this forum more of a dump than it already is

can't even copy-paste correctly. How can you hope to cheat your homework if we don't even know the question? 

Only statements A and C are true.

The factored form is (x - 1)(x - 4)(x + 1)(x + 4).

(a) 56 digits

(b) The probability is 1/10.

(c) The sum is 3320.

n = 14.

That answer is wrong.