Question
What is the pattern rule for 1122, 1112, 1101, 1091?
Answers
oobleck
well, the first three digits are 112,111,110,109, ...
Not sure why the last digit repeats in pairs, but I guess the next two terms could be 1080, 1070
Not sure why the last digit repeats in pairs, but I guess the next two terms could be 1080, 1070
mathhelper
No obvious pattern, but
the difference between first and second is -10
the difference between the 2nd and third is -21
the difference between the 3rd and the 2nd is -11
the difference between the 4th and the 3rd is -10
we could "force" a pattern on the numbers by assuming they are the
results of a cubic function
f(x) = ax^3 + bx^2 + cx + d , and let x = 0,1,2,3
f(0) = 0+0+0+d = 1122 , so d = 1122
f(1) = a + b + c + 1122 = 1112
or a+b+c = -10
f(2) = 8a + 4b + 2c + 1122 = 1101
8a + 4b + 2c = -21
f(3) = 27a + 9b + 3c + 1122 = 1091
27a + 9b + 3c = -31
so term(n) = (1/3)n^3 - (3/2)n^2 - (53/6)n + 1122
for n = 0, 1, 2, 3 to get your 4 terms
check: term(3) should give us 1091
term(3) = (1/3)(27) - (3/2)(9) - (53/6)(3) + 1122
= 9 - 27/2 - 53/2 + 1122
= 1091
term(2) = (1/3)(8) - (3/2)(4) - (53/6)(2) + 1122
= 1101
Looking good
I still believe you have a typo somewhere, so I was just having some fun with this.
the difference between first and second is -10
the difference between the 2nd and third is -21
the difference between the 3rd and the 2nd is -11
the difference between the 4th and the 3rd is -10
we could "force" a pattern on the numbers by assuming they are the
results of a cubic function
f(x) = ax^3 + bx^2 + cx + d , and let x = 0,1,2,3
f(0) = 0+0+0+d = 1122 , so d = 1122
f(1) = a + b + c + 1122 = 1112
or a+b+c = -10
f(2) = 8a + 4b + 2c + 1122 = 1101
8a + 4b + 2c = -21
f(3) = 27a + 9b + 3c + 1122 = 1091
27a + 9b + 3c = -31
so term(n) = (1/3)n^3 - (3/2)n^2 - (53/6)n + 1122
for n = 0, 1, 2, 3 to get your 4 terms
check: term(3) should give us 1091
term(3) = (1/3)(27) - (3/2)(9) - (53/6)(3) + 1122
= 9 - 27/2 - 53/2 + 1122
= 1091
term(2) = (1/3)(8) - (3/2)(4) - (53/6)(2) + 1122
= 1101
Looking good
I still believe you have a typo somewhere, so I was just having some fun with this.