![]() Having this information can also give you peace of mind, knowing that if you were ever in an emergency situation, you could access your belongings quickly. Understanding how to do this can save you time and money, as it means you won’t need to call a locksmith to help you out. It’s not as difficult as it seems to open a stack on a safe without a key, but it does require some knowledge and skill. While hopefully, you’ll never need to break into your safe, the reality is that sometimes you lose the key or forget the combination. Why You Need to Know how To Open a Stack on Safe without A Key So keep on reading to learn more about how to open a stack on safe without a key. In this blog post, we will discuss how you can go about opening a stack on a safe without a key so that you do not need to purchase new keys or hire an expensive locksmith.Īfter reading through our step-by-step guide, even someone who lacks experience with tools and safes should feel comfortable enough to open their own locked stack on the safe quickly and easily. Opening a Stack-On safe without the proper key can seem intimidating however, with some creative problem-solving, you may be able to get your valuables back much sooner than expected. If you wrap the first mul in push dx / pop dx (or just move mov dx, 5 after it), you would get, at the end of the safe, a value in ax equals to 7*(30 +2k) which implies k = 10 indeed.Storing valuable items like jewelry, weapons, important documents, and other treasured possessions in a safe can be a great way to keep them secure, but what happens if you don’t have the key? Mul word ax = 7*(6 + (2k & 0xffff))Ĭonsidering that you have a mov dx, 5 before the first multiplication, did you (or the author of the safe) forget that mul affects dx? I've leaped quite a few logical steps in manipulating the equation and forming the result but, again, thinking of 2k as a shift may help visualize them.Ĭorollary: Due to the algebraic structure involved, this is the only solution. With this condition, the value in ax at the end of the safe can be written as 7*(6 + (2k & 0xffff)) => k & 0x7fff = 22.Īdding the condition stated at the very beginning of this section, the final value for k is 32768 + 22 = 32790 or 0x8016 in hex. If there is a key that unlocks the safe then it must be greater or equal to 32768 so that dx is 1 after the first multiplication. Therefore no key for which dx is zero can unlock the safe.Ī key has dx 0 iif its value is less than 32768 (again, this is easy to see when thinking of the multiplication as a left shift by one).Ĭorollary: 10 cannot be a solution. It is then easy to see that 350 = 7*(5+2k) => 2k = 45 has no solution. ![]() If dx is zero, the whole loopy1 block does nothing (as dx also sets ax) and the value in ax at the end of the safe is 7*(5 +2k) where k is the key (see the commented code below). This multiplication is just doubling the key, so dx is either 0 or 1 (an easy way to see this is to think of the multiplication as a left shift by one or by remembering that the sum of two n-bit numbers has at most n+1 bits) The value in dx used as the counter for the loop instruction comes from the first mul instruction.
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