"I am a 4-digit even number where the sum of my digits is half of the Jersey number made famous by baseball player Jackie Robinson. If you take my second digit and multiply it to my last digit, the result is a 2-digit number - the tens digit in this result is my third digit. My last digit is one more than my third digit and one less than my first digit. If you multiply my third digit to itself, you will get my second digit". What's my number?

I am a 4digit even number where the sum of my digits is half of the Jersey number made famous by baseball player Jackie Robinson If you take my second digit and class=

Respuesta :

Let the number be abcd

As given the sum of the numbers is 21, then [tex]a+b+c+d=21[/tex]

[tex]d=c+1[/tex]

[tex]d=a-1[/tex]  

[tex]b=c^{2}[/tex]

As both 'd' are equal, we have

[tex]a-1=c+1[/tex]

[tex]c=a-2[/tex] and [tex]b=(a-2)^{2}[/tex]

Now, we will substitute for b , c and d in the first equation we get

[tex]a+(a-2)^{2}+a-2+a-1=21[/tex]

[tex]a+a^{2}-4a+4+a-2+a-1=21[/tex]

[tex]a^{2}-a-20=0[/tex]

[tex]a^{2}-5a+4a-20=0[/tex]

[tex]a(a-5)+4(a-5)=0[/tex]

[tex](a+4)(a-5)=0[/tex]

[tex]a=-4 and a=5[/tex]

Neglect the negative value. So a = 5

Hence, the first number is = 5

As d=a-1, so d=5-1 = 4

As c=d-1, so c=4-1=3

As b=[tex]c^{2}[/tex] = [tex]3^{2}=9[/tex]

Hence, the number is 5934 .