let the number be
1abcdef
f=e+d
f=1+b
a=c
a=f
a+b+c+d+e+f+1=45
ok you have five equations and six unknowns. There can be no single solution.
so take the last equation, and start substituting
a + (b+1) + c+ (d+e) + f=45
f+f+f+f+f=45
f=9, a=9, c=9,b=8, or
1989de9 is the number so far...
and d+e=9
so, any combination of d and e such that the combination adds to 9 will work.
0,9
1,8
2,7,
3,4...
lets try 2,7 to see if the number satisfies the stated facts.
1989279 is between one and two million,
1989279 has one digit as the sum of tens and hundreds
1989279 has one digit as the sum of millions and ten thousands digit
1989279 has hundred thousands, thousands, and ones digit the same
1989279 has the sum of digits equal to 45
remember, any combination of d and e adding to 9 will work
I am a number between 1000000 and 2000000. My ones digit is the sum of my tens and hundreds digits and is also the sum of my millions and ten thousands digits.my hundred thousands, thousands, and ones digits are the same.the sum of my digits is 45.what number am i?
1 answer