Average but special numbers
481 = the average of 148, 184, 418, 481, 814 and 841.
What are all the other such 3 digit numbers?
What are all the other such 3 digit numbers?
Labels: mathemagic, mathschallenge
posted by Chris at
2:15 PM
6 Comments:
370, 407, 592, 518, 629 and obviously any number where all three digits are the same.
Hi again. You got them all. How (computer I suppose)?
Average but special
If numbers are in form of a 3 digit number using x,y,z where the average of all 6 combos may be expressed in the form of a 3 digit number using only x,y,z . We solve for x,y,z as follows
x, y, z may only represent numbers from 0 to 9
xyz= 100x +10y +z
xzy= 100x +1y +10z
yzx= 1x +100y + 10z
yxz= 10x +100 y +z
zxy= 10x + 1y + 100z
zyx= 1x + 10y + 100z
Add all and divide by 6 for average we get :
(222x+222y+222z)/6 = 37x+37y+37z
so for the average to equal xyz:
37x+37y+37z=100x+10y+1z
63x-27y-36z=0
7x-3y-4z=0
x can't be 0 if it is the first digit
if x is 1 , y and z must be 1 for (111)
if x is 2, y and z must be 2 for (222)
if x is 3, y and z must be 3 for (333) OR y must be 7 and z must be 0 for (370)
if x is 4,y and z must be 4 for (444) OR y must be 8 and z must be 1 for (481) OR y must be 0 and z must be 7 for(407)
if x is 5,y and z must be 5 for (555) OR y must be 9 and z must be 2 for (592) OR y must be 1 and z must be 8 for (518)
if x is 6, y and z must be 6 for (666) OR y must be 2 and z must be 9 for (629)
if x is 7, y and z must be 7 for (777)
if x is 8, y and z must be 8 for (888)
if x is 9, y and z must be 9 for (999)
ANSWERS:
111, 222,333,444,555,666,777,888,999
370,407, 481, 518,592, 629
Cam
Hi Cam. I can't find an "obvious" quick way of see the whole truth of your 9 if statements. Is there a neat trick, or was it just sweatwork? Obviously yours is the kind of answer I was hoping for. Thanks.
as far as the statements in my solution go, the if x is #, are more of let x be #, than if statements. Simply stated all possible value of x (integers 0 to 9) must be tested to see if they yield valid y and z solutions.
Valid y and z solutions are easily obtained by plugging the value we have assigned for x into 7x and then testing with z=0 to 9. If the remaining quantity is divisible by 3 we have found a valid x,y,z.
e.g. for x=5
7*5=35
We step 35 down by 4s
35,31,(27),23,19,(15),11,7,(3),-1
we identified numbers divisible by 3 (0 or greater and less than or equal to 27 since y must be from 0 to 9)
27/3=9 for y=9, 2 steps for z=2
15/3=5 for y=5, 5 steps for z=5
3/3= 1 for y=1, 8 steps for z=8
It's a fairly quick and dirty method. Not fancy, but quick.
Cam
Hi Cam. Thanks for the feedback.
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