Study guides

Q: 9999 plus 8888 plus 7777 plus 6666 plus 5555 replace 14 digits with 0 so sum comes to 1111?

Write your answer...

Submit

Related questions

9915 1159 5555 9999 1111

0.5555 = 5555/10000 = 1111/2000

5+5+5+1+1+4=21

1, 5, 11, 55, 101, 505, 1111, 5555.

The factors of 5555 are 1, 5, 11, 55, 101, 505, 1111, and 5555. The prime factors of 5555 are 5, 11, and 101.

21

2222= 3 mod 7 5555 = 4 mod 7 2222^5555 + 5555^2222 = (3^5)^1111 + (4^2)^ 1111 as it is a^n + b^n and n is odd so it is divisible by (a+b) 3^5 + 4^ 2 = 259 hence by 7 as 7 is a factor of 259

This is impossible. 21 is an odd number and all these digits are odd. Whenever you add 6 odd numbers together, the answer is always even!

0.05555 = 5555/100000 = 1111/20000

T

There are 18 111, 222, 333, 444, 555, 666, 777, 888, 999, 1111, 2222, 3333, 4444, 5555, 6666, 7777. 8888. 9999

You can make 24 4-digit numbers. 1111 2222 3333 4444 5555 6666 1234 2345 3456 1356 6543 5432 4321 1432 ok i give up!

5555

9/3 + 9+ 5+ 3+ 1

With each digit having only 2 possibilities, the answer is 2 to the 4th power, which is 16. The 4 is because there are 4 digits. Think about the binary numbers 0000 to 1111, there are 2 possibilities for each digit. If your constraint is that the digits must have at least one 2 and at least one 5, then eliminate the two combinations 2222 and 5555, and that answer would be 14.

9/3+9+5+3+1=21

his cd comes out on the 5 of may, 5555

5555 times 5555 is 30,858,025

(7777-6666)/1111 = 1 (4444+5555)/(9999/3333) = 3 8888/2222 = 4 4*3 + 1 =13

5555

5555 + 5555 = 11,110

5555-4444+5555

public static void main(String[] args) { for(int i=1;i<5;i++) { for(int j=1;j<5;j++) { System.out.print(i); } System.out.println(); } }

9+9+9+9=36-5-5-5-5=16+3+3+3-3=22+1-1-1/1=21

They are Numbers xD