How fast does it terminate?

1 x 7 x 8 = 56

5 x 6 = 30

3 x 0 = 0

I believe that for this kind of problem, the simplest solution is the best. However, I would like to propose an alternative solution that satisfies this pattern, including the 'no term could come before 178' condition.

Here is the solution:

Let n denote index of each term. For n = 1: $a_{n}=178$

For n > 1:

$\boxed{a_{n}=\frac{\frac{134n+154}{n-1}-a_{n-1}}{\frac{134n+154}{n-1}-a_{n-1}} \times \frac{a_{n-1}(n-1)-22(7-2n)}{n}}$

For instance:

$a_{2}=\frac{\frac{134(2)+154}{2-1}-178}{\frac{134(2)+154}{2-1}-178} \times \frac{178(2-1)-22(7-2(2))}{2}=56$

$a_{3}=\frac{\frac{134(3)+154}{3-1}-56}{\frac{134(3)+154}{3-1}-56} \times \frac{56(3-1)-22(7-2(3))}{3}=30$

Therefore,

$a_{4}=\frac{\frac{134(4)+154}{4-1}-30}{\frac{134(4)+154}{4-1}-30} \times \frac{30(4-1)-22(7-2(4))}{4}=28$

To sum up, except for 0, the other possible answer is 28 .

Multiply every digit in a number to get next number in sequence.

1x7x8=56,5x6=30 & 3x0=0. So, answer is 0.

see... 1 7 8=56, 5 6=30 and now 3 0= 0... so 0 is the answer

1 * 7 * 8=56

5*6=30

3*0=0

1 x 7 x 8 = 56 then 5 x 6 = 30 next 3 x 0 = 0 so next number is 0.

1 x 7 x 8 = 56 5 x 6 = 30 3 x 0 = 0 so easy

178,56,30,0

1st Term=178

2nd Term=1x7x8=56

3rd Term=5x6=30

4th Term=3x0=0

The heading says it all !!!!

Please change the title !

1 7 8=56 5 6=30 3 0=0

just multiply each number so that you can get the next number. Like 1st number is

178 so

1 *7 *8 =56 = 2nd number

56= 5*6= 30 3rd number

30=3*0= 0 next number

1 *7 *8 =56

5 * 6=30

3 * 0=0

The product of the digits of any number gives the next one

Just have to multiply last two digits of each number