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Find the last two digits of the number 3600 - Mathematics

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प्रश्न

Find the last two digits of the number 3600 

योग

उत्तर

Consider 3600

3600 = (32)300

= 9300

= (10 – 1)300 

(10 – 1)300 = `""^300"C"_0(10)^300 * (- 1)^0 + ""^300"C"_1 (10)^(300 - 1) * (- 1)^1 + .... + ""^300"C"_299 (10)^1 + ""^300"C"_300 (10)^0 (- 1)^300`

= 10300 – 300 (10)299 + ……………. + 300 C1 × 10 × – 1 + 1 × 1 × 1

= 10300 – 300 (10)299 + …………….. – 300 × 10 + 1

= 10300 – 300 × 10299 + …………… – 3000 + 1

All the terms except the last are multiples of 100

And hence divisible by 100.

∴ The last two digits will be 01.

shaalaa.com
Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Binomial Theorem, Sequences and Series - Exercise 5.1 [पृष्ठ २१०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 5 Binomial Theorem, Sequences and Series
Exercise 5.1 | Q 8 | पृष्ठ २१०
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