222a+222b+222c = 222(a+b+c)
a+b+c
=222
What if 2 of the 3 digits are the same?
If 2 digits are the same : -
223 334 566
322 343 656
+232 +433 +665
777 ÷ 7=111 1110 ÷ 10=111 1887 ÷ 17=111
224 559 772
242 595 727
+422 +955 +277
888 ÷ 8=111 2109 ÷19=111 1776 ÷ 16=111
It seems that when 2 of the 3 digits are the same the answer to the problem is 111.
Can I use Algebra to explain this?
aab=100a+10a+b
aba=100a+10b+a
+baa=100b+10a+a
222a+111b = 111(2a+b)
2a+b
=111
What if all 3 digits are the same number?
If All Digits Are The Same : -
333 = 37 444 = 37 666 = 37 999 = 37 777 _ = 37
9 12 18 27 21
It seems that when all the 3 digits are the same number the answer to the problem is 37.
Can I use Algebra to explain this?
aaa=100a+10a+a
=111a = 37
3a
What if I used 4 digits instead 3 digits to investigate the answer to the problem?
If All 4 Digits Are Different : -
1234 2345 3215
1243 2354 3251
1342 2453 3125
1324 2435 3152
1432 2543 3521
1423 2534 3512
2341 3452 2315
2314 3425 2351
2413 3524 2153
2431 3542 2135
2134 3254 2513
2143 3245 2531
3214 4523 1235
3241 4532 1253
3421 4325 1352
3412 4352 1325
3124 4253 1523
3142 4235 1532
4321 5324 5321
4312 5342 5312
4213 5432 5231
4231 5423 5213
4132 _ 5243 _ 5123
+ 4123 = 66660 ÷ 10=6666 +5234 = 93324 ÷ 15=6666| +5132 = 73326 ÷ 11=6666
It seems that for 4 digits, when all digits are different the answer to the problem is 6666.
Can I use Algebra to explain this?
abcd=1000a+100b+10c+d
abdc=1000a+100b+10d+c
adcb=1000a+100d+10c+b
adbc=1000a+100d+10b+c
acbd=1000a+100c+10b+d
acdb=1000a+100c+10d+b
bacd=1000b+100a+10c+d
badc=1000b+100a+10d+c
bdca=1000b+100d+10c+a
bdac=1000b+100d+10a+c
bcda=1000b+100c+10d+a
bcad=1000b+100c+10a+d
cabd=1000c+100a+10b+d
cadb=1000c+100a+10d+b
cbad=1000c+100b+10a+d
cbda=1000c+100b+10d+a
cdba=1000c+100d+10b+a
cdab=1000c+100d+10a+b
dacb=1000d+100a+10c+b
dabc=1000d+100a+10b+c
dbca=1000d+100b+10c+a
dbac=1000d+100b+10a+c
dcab=1000d+100c+10a+b
+dcba=1000d+100c+10b+a
6666a+6666b+6666c+6666d
a+b+c+d
=6666(a+b+c+d) =6666
a+b+c+d
What if 2 of the 4 digits are the same?
If 2 Of The 4 Digits Are The Same : -
1123
1132
1213
1231
1321
1312
2113
2131
2311
3211
3121
+3112
23331 ÷ 7=3333
Is this the same answer every time 2 of the 4 digits are the same?
Let’s try to use Algebraic terms to find out.
aabc=1000a+100a+10b+c
aacb=1000a+100a+10c+b
abac=1000a+100b+10a+c
abca=1000a+100b+10c+a
acab=1000a+100c+10a+b
acba=1000a+100c+10b+a
baac=1000b+100a+10a+c
baca=1000b+100a+10c+a
bcaa=1000b+100c+10a+a
caab=1000c+100a+10a+b
caba=1000c+100a+10b+a
+cbaa=1000c+100b+10a+a
6666a+3333b+3333c
2a+b+c
=3333(2a+b+c) =3333
2a+b+c
What if 3 of the 4 digits are the same?
If 3 Of The 4 Digits Are The Same : -
4447
4474
4744
+7444
21109 ÷19=1111
Is this the same answer every time 3 of the 4 digits are the same?
Lets usr Algebraic terms to find out.
aaab=1000a+100a+10a+b
aaba=1000a+100a+10b+a
abaa=1000a+100b+10a+a
baaa=1000b+100a+10a+a
3333a+1111b
3a+b
=1111(3a+b) =1111
3a+b
What if all 4 digits are the same?
If All 4 Digits Are The Same : -
9999 =277.75
36
Is this the same answer every time all 4 digits are the same?
Let’s use Algebraic terms to find out.
aaaa=1000a+100a+10a+a
=1111a =277.75
4a
What if I do the same problem but with 5 digits instead of 4 or 3?
From the data I have gathered maybe I can guess the outcomes for 5 digits.
5 Digit Number
If All 5 Digits Are Different : -
