Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Math Question?

In order to win the grand prize, paul has to crack this secret code. clues:

a. the third is greater than the second digit

b. the difference between the third and fourth digits is 3

c. when the sum of all numbers is divided by 5, it has a remainder of 3

d. when the sum of all the numbers is divided by 6, it has a remainder of 3​

Relevance
• Alan
Lv 7
1 month ago

1st , you need to clarify if

the difference between the 3 and 4th digits is 3

implies whether you mean

3rd -4th  = 3

or

4th - 3rd = 3

if you don't care , then

so the five digit number

77856

works

the sum = 33

33/6 has remainder 3

33/5 has remainder 3

8> 7

8-5 = 3

5-8 = -3

I first checked for 4 digit number and

it seems impossible.

so if it's a 4 digit number

then  abcd

c> b

c-d =  3

(a+b+c+d)/5  has remainder 3

(a+b +c +d)/6 has remainder 3

since a, b, c, and d are all less than 9

there sum is  36 or less

so 1st sum say one of the following is true

(a+ b+ c+  d) = 8,13,18,23, 33

2nd sum says

(a+b+c +d ) = 9,15, 21,27, 33

What the common value , it is 33

so you know (a+b+c+d) =33

so if 3 of digits were so possible digits include

9,9,9,6

9,9,8,7 (but no two digits are 3 apart)

9,8,8,8  (but no two digits are 3 apart)

9,7,9,8  (but no two digist are 3 apart)

so 9,9,9,6 are a good combination

but no combination of these numbers  can meet the other criteria

so it must a 5 digits number

a+b+c+ d+ (d-3) = 33

or a+ b+ c+d +(d+3) = 33

a+b +c +2d  = 36

a+ b+ c + 2d =30 if d = 4,

then a+ b+ c =  28 (impossible ) or a+ b+ c =  22

then all other digits must be at least 7 and so

could find one worked

but then a saw one the worked if  d = 5

77856

the sum = 33

33/6  has remainder 3

33/5  has remainder 3

8> 7

8-5 = 3

5-8 = -3

• 1 month ago

c > b

c - d = 3

(a + b + c + d )/5 = x + 3

(a + b + c + d)/6 = y + 3