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.

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000

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the last 3 digits of x wiil be 000

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nahh d ans is 450 i cant  find a way to solve dis

by d way how did ya do it?

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