Channel: Hard Question Series💪🏻
Forwarded from 0438/62_Mridul
Last 2 digits of 87^474
@Abhi_Infinity John
@Abhi_Infinity John
Forwarded from °FSociety°
(90 - 3)⁴⁷⁴
Binomial
(474 * 90 * (-3)⁴⁷³ + (-3)⁴⁷⁴ ) mod 100
(6*20 + 69) mod 100
89
Binomial
(474 * 90 * (-3)⁴⁷³ + (-3)⁴⁷⁴ ) mod 100
(6*20 + 69) mod 100
89
Forwarded from °FSociety°
Sum(n*n! )= (n + 1)! - 1
So
(100! - 1) mod 67
-1 mod 67
66
So
(100! - 1) mod 67
-1 mod 67
66
Hard Question Series💪🏻
Last 2 digits of 87^474 @Abhi_Infinity John
Consider only last two digit and reduce its power
87^474 == 69^237 == (09)^79
==(09)*(81^39)
==09 *21
= 89
87^474 == 69^237 == (09)^79
==(09)*(81^39)
==09 *21
= 89
Hard Question Series💪🏻
Last 2 digits of 87^474 @Abhi_Infinity John
87^474
13^474. (87=(100-87)=13)
13^(2*237)
169^237
69^237(consider last two digits)
69*69^236
69*(19^236) ((69=(69-50)=19)
69*(19^2*118)
69*(361^118)
69(61^118)
69*(81) ((abc1^pqr== c*r1))
5589
89(last two digits)
13^474. (87=(100-87)=13)
13^(2*237)
169^237
69^237(consider last two digits)
69*69^236
69*(19^236) ((69=(69-50)=19)
69*(19^2*118)
69*(361^118)
69(61^118)
69*(81) ((abc1^pqr== c*r1))
5589
89(last two digits)
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