153

153 = 3 × 3 × 17

153 = 1 + 23 + 45 + 67 + 8 + 9

153 = 9 + 8 + 76 + 54 + 3 + 2 + 1

153 = 0^5 + 1^8 − 2^7 − 3^9 + 4^2 + 5^6 + 6^3 + 7^0 + 8^4 + 9^1

153 = 1! + 2! + 3! + 4! + 5!

153 = 1³ + 5³ + 3³ (Armstrong number)

153 = 3^2 + 12^2

153 divides 35^2 - 1.

153 = 11 × (11 + 1 + 1 + 1) − 1

153 = 2 × 22 − 2 + 222/2

153 = 3 × (3³ + 3³ − 3)

153 = 4⁴ + 4 + 4 − 444/4

153 = 5 × (5 × 5 + 5) + 5 − (5 + 5)/5

153 = 6 × 6 + 6 + 666/6

153 = 77 + 77 − 7/7

153 = 88 + 8 × 8 + 8/8

153 = 9 × (9 + 9) − 9

a(n) = n*(n+8), n = 9

For any natural number divisible by three continuously calculate the sum of the cubes of the decimal digits: This sequence will always reach 153.

153 = 9 × 17

Number k such that (11*10^k + 19)/3 is prime

10th Hexagonal number (n=9), 2(2n − 1)

153 = binomial(17 + 1, 2) is the 17th triangular number.

Number of distinct products i*j*k for 1 <= i <= j < k <= n, n = 12

Factors: 1, 3, 9, 17, 51, 153

One hundred fifty-three

Representations, Binary to Hexadecimal:

10011001_2
12200_3
2121_4
1103_5
413_6
306_7
231_8
180_9
12a_11
109_12
ba_13
ad_14
a3_15
99_16

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