if 5+55+555+....+555...555=x then the last 3 digits of x will be?
(555...555=100 digits)
plz respond soon
x = 5 + 55 + 555 + ........ +
Sum of digits at the units places
= 5 + 5 + 5 + ...... + 5 (100 terms)
= 5 × 100
= 500
Digits at the unit place of x is 0.
Sum of the digit at the tens places
= 5 + 5 + ...... + 5 (99 terms) + 50 (Carry)
= 5 × 99 + 50
= 495 + 50
= 545
Digit at the tens place of x is 5.
Sum of digits at the hundreds places.
= 5 + 5 + 5 +...... + 5 (98 terms) + 54 (carry)
= 5 × 98 + 54
= 490 + 54
= 544
Digit at the hundreds place of x is 4.
Hence, the last 3-digits of x will be 450.